Trigonometry with cube root

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Trigonometry with cube root

G H J

Reveal topic tags

trig identities

L

G H BBookmark kLocked kLocked NReply

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

149 posts

Aguero

#1 h Apr 20, 2015, 6:37 AM • 4 Y

Y by 62861, jhu08, Adventure10, Mango247

Find $\sqrt [3] {cos \frac{2\pi}{7}}+\sqrt [3] {cos \frac{4\pi}{7}}+\sqrt [3] {cos \frac{8\pi}{7}}$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

965 posts

#2 h Apr 20, 2015, 7:29 AM • 3 Y

Y by jhu08, Adventure10, Mango247

Firstly $\cos{\frac{8\pi}{7}}=\cos{\frac{6\pi}{7}}$ . We know $\cos{\frac{2\pi}{7}}$ , $\cos{\frac{4\pi}{7}}$ , and $\cos{\frac{6\pi}{7}}$ are the roots of the polynomial $8x^3+4x^2-4x-1$ . So let $\sqrt[3]{\cos{\frac{2\pi}{7}}}=a$ , $\sqrt[3]{\cos{\frac{4\pi}{7}}}=b$ , $\sqrt[3]{\cos{\frac{6\pi}{7}}}=c$ , and $a+b+c=x$ . Then cube the expression and do magical stuff.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

2311 posts

#4 h Apr 20, 2015, 7:40 AM • 9 Y

Y by 62861, enhanced, Davrbek, opptoinfinity, Kunihiko_Chikaya, aops29, jhu08, Adventure10, Mango247

This identity is quite famous ( found by Ramanujan )
$\sqrt [3] {\cos \frac{2\pi}{7}}+\sqrt [3] {\cos \frac{4\pi}{7}}+\sqrt [3] {\cos \frac{8\pi}{7}}= \sqrt [3]{\frac{5-3\sqrt [3]{7}}{2}}$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1510 posts

mathuz

#5 h Apr 20, 2015, 7:54 AM • 3 Y

Y by jhu08, Adventure10, Mango247

very nice who can more informations...

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

2311 posts

#7 h Apr 20, 2015, 10:17 AM • 10 Y

Y by Tintarn, 62861, lebathanh, enhanced, opptoinfinity, Centralorbit, Pluto1708, aops29, jhu08, Adventure10

My solution :

Let $\alpha= \sqrt [3] {\cos \frac{2\pi}{7}}, \beta=\sqrt [3] {\cos \frac{4\pi}{7}}, \gamma=\sqrt [3] {\cos \frac{8\pi}{7}}$ , so $(x-{\alpha}^3)(x-{\beta}^3)(x-{\gamma}^3)=8x^3+4x^2-4x-1$ ,

hence we get (1) ${\alpha}^3+{\beta}^3+{\gamma}^3=\frac{-1}{2}$ , (2) ${\alpha}^3{\beta}^3+ {\beta}^3{\gamma}^3+ {\gamma}^3{\alpha}^3=\frac{-1}{2}$ , (3) ${\alpha}^3{\beta}^3{\gamma}^3=\frac{1}{8} \Longrightarrow {\alpha}{\beta}{\gamma}=\frac{1}{2}$ .

Let $\mathcal{P}={\alpha}+{\beta}+{\gamma}$ and $\mathcal{Q}={\alpha}{\beta}+ {\beta}{\gamma}+ {\gamma}{\alpha}$ .

From $-2={\alpha}^3+{\beta}^3+{\gamma}^3-3{\alpha}{\beta}{\gamma}=\mathcal{P}(\mathcal{P} ^2-3\mathcal{Q} ) \Longrightarrow$ (4) $\mathcal{P} ^3+2=3 \mathcal{P} \mathcal{Q}$ .

From (1), (3) and $\sum ( {\beta} ^2 {\gamma}+{\beta} {\gamma} ^2) =\mathcal{P} (\mathcal{P} ^2- 2 \mathcal{Q} )- \sum {\alpha}^3$

we get $\mathcal{Q} ^3=\sum {\beta}^3 {\gamma} ^3+6 {\alpha}^2 {\beta}^2 {\gamma}^2 +\frac{3}{2} ( \mathcal{P} (\mathcal{P} ^2- 2 \mathcal{Q} )+\frac{1}{2})$ ,

so combine with (4) we get $\mathcal{Q} ^3=\frac{7}{4}+\frac{3}{2} \mathcal{P} ^3-3 \mathcal{P} \mathcal{Q} =\frac{-1}{4}+\frac{1}{2} \mathcal{P} ^3 \Longrightarrow$ (5) $\mathcal{Q} ^3=\frac{1}{4} ( 2 \mathcal{P} ^3 -1)$ .

From (4), (5) $\Longrightarrow ( \mathcal{P} ^3 +2)^3=27 \mathcal{P} ^3 \mathcal{Q} ^3=\frac{27}{4} \mathcal{P} ^3 ( 2 \mathcal{P} ^3 -1 ) \Longrightarrow 8 (\mathcal{P} ^3 )^3-60 (\mathcal{P} ^3)^2+150 \mathcal{P} ^3+64=0$ ,

hence we get $(2\mathcal{P} ^3 -5)^3 = -189 \Longrightarrow \mathcal{P} ^3=\frac{5-3\sqrt [3]{7}}{2} \Longrightarrow \sqrt [3] {\cos \frac{2\pi}{7}}+\sqrt [3] {\cos \frac{4\pi}{7}}+\sqrt [3] {\cos \frac{8\pi}{7}}=\mathcal{P}= \sqrt [3]{\frac{5-3\sqrt [3]{7}}{2}}$ .

Q.E.D

This post has been edited 1 time. Last edited by TelvCohl, Apr 20, 2015, 3:06 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

23442 posts

pco

#8 h Apr 20, 2015, 3:01 PM • 14 Y

Y by drmzjoseph, Tintarn, YadisBeles, 62861, nicky-glass, opptoinfinity, Hedy, Mateescu Constantin, Hermitianism, teomihai, jhu08, MathLover_ZJ, Adventure10, Mango247

On a more general case, we can establish that, if $a,b,c$ are roots of $x^3+ux^2+(u-3)x-1=0$ , then :

$\sqrt[3]a+\sqrt[3]b+\sqrt[3]c$ $=\sqrt[3]{6-u-3\sqrt[3]{u^2-3u+9}}$

Some examples :

1) $u=1$ gives :
$\sqrt[3]{\cos\frac{2\pi}7}$ $+\sqrt[3]{\cos\frac{4\pi}7}$ $+\sqrt[3]{\cos\frac{6\pi}7}$ $=\sqrt[3]{\frac{5-3\sqrt[3]7}2}$

2) $u=0$ gives :
$\sqrt[3]{\cos\frac{\pi}9}$ $+\sqrt[3]{\cos\frac{7\pi}9}$ $+\sqrt[3]{\cos\frac{13\pi}9}$ $=\sqrt[3]{\frac{6-3\sqrt[3]9}2}$

3) $u=3$ gives
$\frac 1{\sqrt[3]{\cos\frac{\pi}9}}$ $+\frac 1{\sqrt[3]{\cos\frac{7\pi}9}}$ $+\frac 1{\sqrt[3]{\cos\frac{13\pi}9}}$ $=\sqrt[3]{6-6\sqrt[3]9}$
or also
$\sqrt[3]{2\cos\frac{4\pi}9-1}$ $+\sqrt[3]{2\cos\frac{10\pi}9-1}$ $+\sqrt[3]{2\cos\frac{16\pi}9-1}$ $=\sqrt[3]{3-3\sqrt[3]9}$

4) $u=-3$ gives
$\sqrt[3]{2\sqrt 3\cos\frac{\pi}{18}+1}$ $+\sqrt[3]{2\sqrt 3\cos\frac{13\pi}{18}+1}$ $+\sqrt[3]{2\sqrt 3\cos\frac{25\pi}{18}+1}$ $=0$

And likely some other funny results ...

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

30148 posts

arqady

#9 h Jul 18, 2016, 12:25 PM • 5 Y

Y by 62861, aops29, jhu08, Adventure10, Mango247

I got the same result, but in another form.
Let $a\in\mathbb R$ and $\phi=\arctan\frac{2a+3}{3\sqrt3}$ . Prove that:
$\sqrt[3]{\frac{\sqrt3}{2}\tan\frac{\phi}{3}-\frac{1}{2}}+\sqrt[3]{\frac{\sqrt3}{2}\tan\frac{\phi+\pi}{3}-\frac{1}{2}}+\sqrt[3]{\frac{\sqrt3}{2}\tan\frac{\phi+2\pi}{3}-\frac{1}{2}}=\sqrt[3]{a+6-3\sqrt[3]{a^2+3a+9}}$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

30148 posts

arqady

#10 h Jul 18, 2016, 12:38 PM • 4 Y

Y by 62861, galav, jhu08, Adventure10

More problems in the Ramanujan's style.
1.Prove that $\sqrt[3]{21\sqrt[3]6-37}=\sqrt[3]2+\sqrt[3]9-\sqrt[3]{12}$
2. Prove that $\sqrt[3]{21\sqrt6-17}=\sqrt[3]{18}+\sqrt[3]4-\sqrt[3]3$
3. Solve the following equation. $\sqrt{4+\sqrt{4+\sqrt{4-x}}}=x$
4. Let $a$ , $b$ and $c$ be roots of the equation $x^3-16x^2-57x+1=0$ . Prove that:
$\sqrt[5]a+\sqrt[5]b+\sqrt[5]c=1$ 5. Let $a$ , $b$ and $c$ be roots of the equation $x^3+15x^2-198x+1=0$ . Prove that:
$\sqrt[5]a+\sqrt[5]b+\sqrt[5]c=0$ 6. Solve the following equation.
$x^5+x^4-12x^3-21x^2+x+5=0$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

algebra_star1234

2467 posts

algebra_star1234

#11 h Jul 18, 2016, 12:42 PM • 3 Y

Y by jhu08, Adventure10, Mango247

How does one get the equation $8x^3+4x^2-4x-1$ ?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

30148 posts

arqady

#12 h Jul 18, 2016, 12:47 PM • 3 Y

Y by jhu08, Adventure10, Mango247

It's the Viète's theorem.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

algebra_star1234

2467 posts

algebra_star1234

#13 h Jul 18, 2016, 2:12 PM • 3 Y

Y by jhu08, Adventure10, Mango247

I know these formulas but how to calculate the sum as 4.... (Sorry if this is some trivial fact or something)

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

30148 posts

arqady

#14 h Jul 18, 2016, 4:31 PM • 6 Y

Y by DavidDavid, 62861, algebra_star1234, FUSK, jhu08, Adventure10

For example, $\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}\cos\frac{8\pi}{7}=\frac{8\sin\frac{2\pi}{7}\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}\cos\frac{8\pi}{7}}{8\sin\frac{2\pi}{7}}=\frac{\sin\frac{16\pi}{7}}{8\sin\frac{2\pi}{7}}=\frac{1}{8}$ and
$\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{8\pi}{7}=\frac{2\sin\frac{\pi}{7}\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right)}{2\sin\frac{\pi}{7}}=$
$=\frac{\sin\frac{3\pi}{7}-\sin\frac{\pi}{7}+\sin\frac{5\pi}{7}-\sin\frac{3\pi}{7}+\sin\frac{7\pi}{7}-\sin\frac{5\pi}{7}}{2\sin\frac{\pi}{7}}=-\frac{1}{2}$ .

This post has been edited 1 time. Last edited by arqady, Jul 18, 2016, 4:37 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

41020 posts

sqing

#16 h Oct 13, 2018, 2:51 AM • 3 Y

Y by jhu08, Adventure10, Mango247

pco wrote:

On a more general case, we can establish that, if $a,b,c$ are roots of $x^3+ux^2+(u-3)x-1=0$ , then :

$\sqrt[3]a+\sqrt[3]b+\sqrt[3]c$ $=\sqrt[3]{6-u-3\sqrt[3]{u^2-3u+9}}$

Some examples :

1) $u=1$ gives :
$\sqrt[3]{\cos\frac{2\pi}7}$ $+\sqrt[3]{\cos\frac{4\pi}7}$ $+\sqrt[3]{\cos\frac{6\pi}7}$ $=\sqrt[3]{\frac{5-3\sqrt[3]7}2}$

2) $u=0$ gives :
$\sqrt[3]{\cos\frac{\pi}9}$ $+\sqrt[3]{\cos\frac{7\pi}9}$ $+\sqrt[3]{\cos\frac{13\pi}9}$ $=\sqrt[3]{\frac{6-3\sqrt[3]9}2}$

3) $u=3$ gives
$\frac 1{\sqrt[3]{\cos\frac{\pi}9}}$ $+\frac 1{\sqrt[3]{\cos\frac{7\pi}9}}$ $+\frac 1{\sqrt[3]{\cos\frac{13\pi}9}}$ $=\sqrt[3]{6-6\sqrt[3]9}$
or also
$\sqrt[3]{2\cos\frac{4\pi}9-1}$ $+\sqrt[3]{2\cos\frac{10\pi}9-1}$ $+\sqrt[3]{2\cos\frac{16\pi}9-1}$ $=\sqrt[3]{3-3\sqrt[3]9}$

4) $u=-3$ gives
$\sqrt[3]{2\sqrt 3\cos\frac{\pi}{18}+1}$ $+\sqrt[3]{2\sqrt 3\cos\frac{13\pi}{18}+1}$ $+\sqrt[3]{2\sqrt 3\cos\frac{25\pi}{18}+1}$ $=0$

And likely some other funny results ...

https://mp.weixin.qq.com/s/kkSUqkq3it7Ekbsayea6JA

https://artofproblemsolving.com/community/c6h346917p1858263:
$\cot^2 \frac{\pi}{7}+\cot^2 \frac{2\pi}{7}+\cot^2 \frac{3\pi}{7}=5$ https://artofproblemsolving.com/community/c6h1507047p8925730:
$\sqrt[3]{\cos\frac{2\pi}{7}}+\sqrt[3]{\cos\frac{4\pi}{7}}+\sqrt[3]{\cos\frac{6\pi}{7}}=\sqrt[3]{\frac{5-3\sqrt[3]{7}}{2}}$

This post has been edited 2 times. Last edited by sqing, Oct 30, 2018, 2:12 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

41020 posts

sqing

#17 h Dec 31, 2018, 2:18 AM • 3 Y

Y by jhu08, Adventure10, Mango247

Prove that:
$cot\frac{\pi}{26}-4 sin\frac{5\pi}{13}=\sqrt{13+2\sqrt{13}}$ (Vasile Mircea Popa)

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

41020 posts

sqing

#18 h Jan 10, 2019, 2:53 AM • 2 Y

Y by jhu08, Adventure10

Prove that:
$csc\frac{\pi}{14}-4 cos\frac{2\pi}{7}=2.$ $\csc{\frac{\pi}{10}}-4 \cos{\frac{2 \pi}{5}}=2$ $\csc{\frac{5\pi}{14}}-4 \cos{\frac{10\pi}{7}}=2$ VMP

Attachments:

This post has been edited 4 times. Last edited by sqing, Jan 11, 2019, 2:37 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

41020 posts

sqing

#19 h Jan 10, 2019, 4:24 AM • 3 Y

Y by jhu08, Adventure10, Mango247

$\tan\frac{\pi}{7} \tan\frac{2\pi}{7} \tan\frac{3\pi}{7}=\sqrt7$ https://artofproblemsolving.com/community/c129h1359338p7479664
$\cos \frac{2\pi}{13} \cdot \cos \frac{6\pi}{13} \cdot \cos \frac{8\pi}{13} = \frac{3 - \sqrt{13}}{16}$ https://artofproblemsolving.com/community/c7h1206420p5965720
$\sin\frac{\pi}{7}\sin\frac{3\pi}{7}\sin\frac{5\pi}{7}=\frac{\sqrt{7}}{8}$ $\tan^2 20^\circ + \tan^2 40^\circ + \tan^2 80^\circ = 33$ https://artofproblemsolving.com/community/c6h1870830p12690838

Prove that(Ahmed Salama Hegazy):

Attachments:

This post has been edited 3 times. Last edited by sqing, Jul 9, 2019, 1:08 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

108 posts

#24 h Jun 6, 2019, 1:25 PM • 3 Y

Y by aops29, jhu08, Adventure10

arqady wrote:

Let $a$ , $b$ and $c$ be roots of the equation $x^3-16x^2-57x+1=0$ . Prove that:
$\sqrt[5]a+\sqrt[5]b+\sqrt[5]c=1$

arqady, I really want to know your solution, because my is very ugly and done by software.
$a\rightarrow a^5$ , $b\rightarrow b^5$ , $c\rightarrow c^5$ .
Let $3u=a+b+c$ , $3v^2=ab+bc+ca$ ( $v^2$ can be negative) and $w^3=abc$ .
$\begin{cases}243u^5-405u^3v^2+135uv^4+(45u^2-15v^2)w^3=16\\243v^{10}-405uv^6w^3+135u^2v^2w^6+45v^4w^6-15uw^9=-57\\w^{15}=-1\end{cases}$
Since $w^3=-1$ , then
$135uv^4-(405u^3-15)v^2+243u^5-45u^2-16=0$
$v^2=\frac{135u^3\pm\sqrt{5(729u^6+270u^3+192u+5)}}{90u}$
For clearance $x=v^2$ .
Case $x=\frac{135u^3-\sqrt{5(729u^6+270u^3+192u+5)}}{90u}$ , by maxima

expand(subst ([x=(135*u^3-5-sqrt(5*(729*u^6+270*u^3+192*u+5)))/(90*u)], 81*x^5+135*u*x^3+15*x^2+45*u^2*x+5*u+19));
ratsimp(%);

Gives $14348907\sqrt{5^3}u^{15}+7440174\sqrt{5^3}u^{12}+2834352\sqrt{5^3}u^{10}+39366\sqrt{5^7}u^9+ 419904\sqrt{5^3}u^7+4374\sqrt{5^5}u^6+200718\sqrt{5^3}u^5-2304\sqrt{5^3}u^2-48\sqrt{5^5}u-\sqrt{5^5}=(5845851u^{12}+1869885u^9+454896u^7+98415u^6+6480u^4- 675u^3+2304u^2+720u+25)\sqrt{729u^6+270u^3+192u+5}$
squarring
$(14348907\sqrt{5^3}u^{15}+7440174\sqrt{5^3}u^{12}+2834352\sqrt{5^3}u^{10}+39366\sqrt{5^7}u^9+ 419904\sqrt{5^3}u^7+4374\sqrt{5^5}u^6+200718\sqrt{5^3}u^5-2304\sqrt{5^3}u^2-48\sqrt{5^5}u-\sqrt{5^5})^2=(5845851u^{12}+1869885u^9+454896u^7+98415u^6+6480u^4- 675u^3+2304u^2+720u+25)^2(729u^6+270u^3+192u+5)$
$LHS-RHS$
$823564528378596u^{30}+1525119496997400u^{27}-271132355021760u^{25}+988503377683500u^{24}-150629086123200u^{22}+ 282429536481000u^{21}+532571449408740u^{20}-9298091736000u^{19}+35836395232500u^{18}+247716660666600u^{17}+4476858984000u^{16}+ 97373978727120u^{15}+37751974317000u^{14}-1211685480000u^{13}+16083902432700u^{12}-671209983000u^{11}+4516257657780u^{10}-486189783000 u^9-75582720000u^8-126279129600u^7-14328130500u^6-1347873372u^5=972u^5(3u-1)(531441u^{12}+531441u^{11}+354294u^{10}+688905u^9+885735u^8+509571u^7+ 94041u^6+96714u^5+106515u^4+36180u^3+10044u^2+1029u+71)(531441u^{12}- 354294u^{11}+59049u^{10}+492075u^9-623295u^8+236196u^7+305451u^6-360126u^5+ 336150u^4-178335u^3+109719u^2-16851u+19531)=0$ . Obviously $u=0$ is not solution.

allroots(531441*u^12+531441*u^11+354294*u^10+688905*u^9+885735*u^8+509571*u^7+
94041*u^6+96714*u^5+106515*u^4+36180*u^3+10044*u^2+1029*u+71);

Gives

[u=0.08873583372698096*%i-0.06285164678937094,u=-0.08873583372698096*%i-0.06285164678937094,
u=0.249818724727736*%i-0.1038150198772957,
u=-0.249818724727736*%i-.1038150198772957,u=0.4141258170774749*%i+0.40187021328863,
u=0.40187021328863-0.4141258170774749*%i,u=0.04493113182149037*%i-0.7343941915363165,
u=-0.04493113182149037*%i-0.7343941915363165,u=0.6072801786518509*%i-0.5685368799552993,
u=-0.6072801786518509*%i-.5685368799552993,u=0.9566536050241474*%i+0.5677275248696524,
u=0.5677275248696524-0.9566536050241474*%i]

allroots(531441*u^12-354294*u^11+59049*u^10+492075*u^9-623295*u^8+236196*u^7+305451*u^6-360126*u^5+
336150*u^4-178335*u^3+109719*u^2-16851*u+19531);

Gives

[u = 0.4301615523642846 %i - 0.1842082400516285, 
u = (- 0.4301615523642846 %i) - 0.1842082400516285, 
u = 0.5973695492100051 %i - 0.001309564037543536, 
u = (- 0.5973695492100051 %i) - 0.001309564037543536, 
u = 0.5364026821017057 %i + 0.5180417823368586, 
u = 0.5180417823368586 - 0.5364026821017057 %i, 
u = 0.3315150891954592 %i + 0.81690633083384, 
u = 0.81690633083384 - 0.3315150891954592 %i, 
u = 0.9665642344659907 %i + 0.2055111206314012, 
u = 0.2055111206314012 - 0.9665642344659907 %i, 
u = 0.2658544600145461 %i - 1.021608096379594, 
u = (- 0.2658544600145461 %i) - 1.021608096379594]

Case $x=\frac{135u^3-5+\sqrt{5(729u^6+270u^3+192u+5)}}{90u}$ , since $u^2\geq v^2$ , then $u\leq -1.15804$
by maxima

ratsimp(ratsimp(expand(subst ([x=(135*u^3-5+sqrt(5*(729*u^6+270*u^3+192*u+5)))/(90*u)], 81*x^5+135*u*x^3+15*x^2+45*u^2*x+5*u+19))));

$14348907\sqrt{5^3}u^{15}+7440174\sqrt{5^3}u^12+2834352\sqrt{5^3}u^10+39366\sqrt{5^7}u^9 +419904\sqrt{5^3}u^7+4374\sqrt{5^5}u^6+200718\sqrt{5^3}u^5-2304\sqrt{5^3}u^2-48\sqrt{5^5}u-\sqrt{5^5}=-\sqrt{729u^6+270u^3+192u+5} (5845851u^{12}+1869885u^9+454896u^7+98415u^6+6480u^4-675u^3+2304u^2+720u+25)$
Squrring
$(14348907\sqrt{5^3}u^{15}+7440174\sqrt{5^3}u^12+2834352\sqrt{5^3}u^10+39366\sqrt{5^7}u^9 +419904\sqrt{5^3}u^7+4374\sqrt{5^5}u^6+200718\sqrt{5^3}u^5-2304\sqrt{5^3}u^2-48\sqrt{5^5}u-\sqrt{5^5})^2=(5845851u^{12}+1869885u^9+454896u^7+98415u^6+6480u^4-675u^3+2304u^2+720u+25)^2(729u^6+270u^3+192u+5)$
$LHS-RHS$
$-24912826983452529u^{30}-25164471700457100u^{27}-10438595668337760u^{25}-2541865828329000u^{24}-1656919947355200u^{22}+ 203976887458500u^{21}-187449529397760u^{20}-9298091736000u^{19}+35836395232500u^{18}+255981631098600u^{17}+5337793404000u^{16}+ 97391914860870u^{15}+37751974317000u^{14}-1211685480000u^{13}+26705675099887200u^{12}-671209983000u^{11}+10171979570973780u^{10}+ 3529883016229500u^9-75582720000u^8+1506164582102400u^7+78438320892000u^6+720019630933128u^5-8264970432000u^2-860934420000 u+25736373575697375=0$
Again typing in maxima

solve(-24912826983452529*u^30-25164471700457100*u^27-10438595668337760*u^25-2541865828329000*u^24-1656919947355200*u^22+
203976887458500*u^21-187449529397760*u^20-9298091736000*u^19+35836395232500*u^18+255981631098600*u^17+5337793404000*u^16+
97391914860870*u^15+37751974317000*u^14-1211685480000*u^13+26705675099887200*u^12-671209983000*u^11+10171979570973780*u^10+
3529883016229500*u^9-75582720000*u^8+1506164582102400*u^7+78438320892000*u^6+720019630933128*u^5-8264970432000*u^2-860934420000*
u+25736373575697375,u);

Gives

[[0=102521921742603*u^30+103557496709700*u^27+42957183820320*u^25+10460353203000*u^24+6818600606400*u^22-839411059500*
u^21+771397240320*u^20+38263752000*u^19-147474877500*u^18-1053422350200*u^17-21966228000*u^16-400789773090*u^15-155357919000*u^14+
4986360000*u^13-109899897530400*u^12+2762181000*u^11-41859998234460*u^10-14526267556500*u^9+311040000*u^8-6198208156800*u^7-
322791444000*u^6-2963043748696*u^5+34012224000*u^2+3542940000*u-105911002369125]=0]

typing

allroots(102521921742603*u^30+103557496709700*u^27+42957183820320*u^25+10460353203000*u^24+6818600606400*u^22-839411059500*
u^21+771397240320*u^20+38263752000*u^19-147474877500*u^18-1053422350200*u^17-21966228000*u^16-400789773090*u^15-155357919000*u^14+
4986360000*u^13-109899897530400*u^12+2762181000*u^11-41859998234460*u^10-14526267556500*u^9+311040000*u^8-6198208156800*u^7-
322791444000*u^6-2963043748696*u^5+34012224000*u^2+3542940000*u-105911002369125,u);

gives

u = 0.6701030312889549 - 0.7418603072941804 %i, 
u = 0.6936763996501353 %i - 0.6394643083013833, 
u = (- 0.6936763996501353 %i) - 0.6394643083013833, 
u = 0.8693833939058644 %i - 0.5003277841706911, 
u = (- 0.8693833939058644 %i) - 0.5003277841706911, u = 1.003024944024268, 
u = 0.9515836439685664 %i + 0.308918723378388, 
u = 0.308918723378388 - 0.9515836439685664 %i, 
u = 0.5896639595395451 %i - 0.8091698531500543, 
u = (- 0.5896639595395451 %i) - 0.8091698531500543, 
u = 0.9340675629015904 %i - 0.2681691050314659, 
u = (- 0.9340675629015904 %i) - 0.2681691050314659, 
u = 0.9930125376811316 %i + 0.1466433765516465, 
u = 0.1466433765516465 - 0.9930125376811316 %i, 
u = 0.9950264234389671 %i - 0.1076485022948794, 
u = (- 0.9950264234389671 %i) - 0.1076485022948794, 
u = 0.3220753007654913 %i - 0.9341384895486041, 
u = (- 0.3220753007654913 %i) - 0.9341384895486041, 
u = 0.2386333647302399 %i + 0.9080121402521805, 
u = 0.9080121402521805 - 0.2386333647302399 %i, u = - 1.132184434639223, 
u = 0.9072105915610396 %i + 0.5644742490679983, 
u = 0.5644742490679983 - 0.9072105915610396 %i, 
u = 0.6763316638574209 %i + 0.7886132313994743, 
u = 0.7886132313994743 - 0.6763316638574209 %i, 
u = 0.2073934966713666 %i - 0.9781321310923142, 
u = (- 0.2073934966713666 %i) - 0.9781321310923142, 
u = 0.406721388877138 %i + 0.9148651669582274, 
u = 0.9148651669582274 - 0.406721388877138 %i]

So only $u=\frac{1}{3}$ is real solution

This post has been edited 5 times. Last edited by r1chard_castle, Jun 7, 2019, 8:03 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

108 posts

#26 h Jun 10, 2019, 5:56 AM • 3 Y

Y by jhu08, Adventure10, Mango247

r1chard_castle wrote:

arqady, I really want to know your solution, because my is very ugly and done by software.

Did you use trigonometry or like me $uvw$

.

This post has been edited 3 times. Last edited by r1chard_castle, Jun 10, 2019, 6:04 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

108 posts

#27 h Jun 10, 2019, 12:07 PM • 3 Y

Y by jhu08, Adventure10, Mango247

So trigonometry or something else? Give a hint at least.

This post has been edited 1 time. Last edited by r1chard_castle, Jun 10, 2019, 12:25 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

23442 posts

pco

#28 h Jun 11, 2019, 7:15 AM • 3 Y

Y by jhu08, Adventure10, Mango247

I was asked in PM to prove this :

pco wrote:

On a more general case, we can establish that, if $a,b,c$ are roots of $x^3+ux^2+(u-3)x-1=0$ , then :
$\sqrt[3]a+\sqrt[3]b+\sqrt[3]c$ $=\sqrt[3]{6-u-3\sqrt[3]{u^2-3u+9}}$ .

Here is one proof (maybe not the best) :

Let $a,b,c$ be root of $P(x)=x^3+ux^2+(u-3)x-1=0$
So, let $\sqrt[3]a,\sqrt[3]b,\sqrt[3]c$ be root of $Q(x)=x^3-vx^2+wx-1=0$
Let $j=e^{\frac{2i\pi}3}$

Using the fact that $Q(x)Q(jx)Q(j^2x)=P(x^3)$ , we easily get
$v^3-3vw=-u-3$
$w^3-3vw=u-6$

Adding, we get $v^3+w^3=6vw-9$
Multiplying, we get $(w^3-3vw)(v^3-3vw)=(-3-u)(u-6)=-u^2+3u+18$
Which is $v^3w^3-3vw(v^3+w^3)+9v^2w^2=-u^2+3u+18$
And so $v^3w^3-3vw(6vw-9)+9v^2w^2=-u^2+3u+18$
Which is $v^3w^3-9v^2w^2+27vw=-u^2+3u+18$
Or also $(vw-3)^3=-u^2+3u-9$

And so $vw=3-\sqrt[3]{u^2-3u+9}$
implies $3vw-3-u=6-u-3\sqrt[3]{u^2-3u+9}$

And so $v^3=6-u-3\sqrt[3]{u^2-3u+9}$

Which is $\boxed{v=\sqrt[3]{6-u-3\sqrt[3]{u^2-3u+9}}}$

Q.E.D.

This post has been edited 1 time. Last edited by pco, Jun 11, 2019, 7:16 AM
Reason: Typo

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

30148 posts

arqady

#29 h Jun 11, 2019, 6:14 PM • 3 Y

Y by jhu08, Adventure10, Mango247

r1chard_castle wrote:

So trigonometry or something else? Give a hint at least.

I used trigonometry.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

MathPassionForever

1663 posts

MathPassionForever

#30 h Jun 11, 2019, 6:34 PM • 3 Y

Y by jhu08, Adventure10, Mango247

Hey guys! Needed a little help here. I saw these two equalities which are very similar:
$\tan\left(\frac{3\pi}{7}\right)-4\sin \left(\frac{\pi}{7}\right)=\sqrt{7}$ $\tan\left(\frac{3\pi}{11}\right)+4\sin \left(\frac{2\pi}{11}\right)=\sqrt{11}$ Can we generalise these? I had proved just the second one, and it involved squaring the thing and converting every T-ratio to sine function. I found some five sines with angles in an $\text{AP}$ which added nicely to $-1$ . That might be of help I guess

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

41020 posts

sqing

#31 h Jun 11, 2019, 10:14 PM • 2 Y

Y by jhu08, Adventure10

MathPassionForever wrote:

$\tan\left(\frac{3\pi}{7}\right)-4\sin \left(\frac{\pi}{7}\right)=\sqrt{7}$ $\tan\left(\frac{3\pi}{11}\right)+4\sin \left(\frac{2\pi}{11}\right)=\sqrt{11}$

https://artofproblemsolving.com/community/c6h190479p1046116
https://artofproblemsolving.com/community/c4h532603p3086397
https://artofproblemsolving.com/community/c6h562893p3285458
https://artofproblemsolving.com/community/c6h626749p3759491

This post has been edited 1 time. Last edited by sqing, Jun 11, 2019, 10:18 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

108 posts

#32 h Jul 3, 2019, 8:27 AM • 3 Y

Y by jhu08, Master_of_Aops, Adventure10

arqady wrote:

4. Let $a$ , $b$ and $c$ be roots of the equation $x^3-16x^2-57x+1=0$ . Prove that:
$\sqrt[5]a+\sqrt[5]b+\sqrt[5]c=1$

Let $f(x)=(x-a)(x-b)(x-c)=x^3-16x^2-57x+1$
$f(t^5)=t^{15}-16x^{10}-57x+1=(t^3-t^2-2t+1)(t^{12}+t^{11}+3t^{10}+4t^9+9t^8-2t^7 +12t^6-t^5+25t^4+11t^3+5t^2+2t+1)$
Since $f(t^5)$ has only three real roots and since all roots of $t^3-t^2-2t+1$ are reals, then by Vieta's theorem $t_1+t_2+t_3=\sqrt[5]{a}+\sqrt[5]{b}+\sqrt[5]{c}=1$

arqady wrote:

5. Let $a$ , $b$ and $c$ be roots of the equation $x^3+15x^2-198x+1=0$ . Prove that:
$\sqrt[5]a+\sqrt[5]b+\sqrt[5]c=0$

Let $f(x)=(x-a)(x-b)(x-c)=x^3+15x^2-198x+1$
$f(t^5)=t^{15}+15t^{10}-198t^5+1=(t^3-3t+1)(t^{12}+3t^{10}-t^9+9t^8+9t^7+28t^6+ 18t^5+75t^4+26t^3+9t^2+3t+1)$
Since $f(t^5)$ has only three real roots and since all roots of $t^3-3t+1$ are reals, then by Vieta's theorem $t_1+t_2+t_3=\sqrt[5]a+\sqrt[5]b+\sqrt[5]c=0$

This post has been edited 2 times. Last edited by r1chard_castle, Jul 3, 2019, 1:09 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

2944 posts

#33 h May 10, 2020, 1:53 PM • 3 Y

Y by Math-wiz, aops29, jhu08

memorabile posts!

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

41020 posts

sqing

#34 h May 14, 2020, 8:38 AM • 2 Y

Y by teomihai, jhu08

4 4
4
Note that :
1) $x^{15}-16x^{10}-57x^5+1$ $=(x^3-x^2-2x+1)(x^{12}+x^{11}+3x^{10}+4x^9+9x^8-2x^7+12x^6-x^5+25x^4+11x^3+5x^2+2x+1)$

2) $x^3-x^2-2x+1=0$ has three distinct real roots $u,v,w$

So $u^5,v^5,w^5$ are three distinct real roots of $x^3-16x^2-57x+1$ and so are $a,b,c$

And since $u+v+w=1$ , we got the result.
5
Note that $x^{15}+15x^{10}-198x^5+1=(x^3-3x+1)(x^{12}+3x^{10}-x^9+9x^8+9x^7+28x^6+18x^5+75x^4+26x^3+9x^2+3x+1)$
And so the three distinct real roots $x_1=-2\cos\frac{\pi}9$ , $x_2=-2\cos\frac{7\pi}9$ and $x_3=-2\cos\frac{13\pi}9$ of the cubic $x^3-3x+1=0$ are such that :
$x_1^5, x_2^5, x_3^5$ are three distinct real roots of $x^3+15x^2-198x+1=0$
$x_1+x_2+x_3=0$
Q.E.D.

This post has been edited 4 times. Last edited by sqing, May 14, 2020, 8:53 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

41020 posts

sqing

#35 h Jul 25, 2021, 2:59 PM • 1 Y

Y by jhu08

pco wrote:

On a more general case, we can establish that, if $a,b,c$ are roots of $x^3+ux^2+(u-3)x-1=0$ , then :

$\sqrt[3]a+\sqrt[3]b+\sqrt[3]c$ $=\sqrt[3]{6-u-3\sqrt[3]{u^2-3u+9}}$

Some examples :
3) $u=3$ gives
$\frac 1{\sqrt[3]{\cos\frac{\pi}9}}$ $+\frac 1{\sqrt[3]{\cos\frac{7\pi}9}}$ $+\frac 1{\sqrt[3]{\cos\frac{13\pi}9}}$ $=\sqrt[3]{6-6\sqrt[3]9}$
or also
$\sqrt[3]{2\cos\frac{4\pi}9-1}$ $+\sqrt[3]{2\cos\frac{10\pi}9-1}$ $+\sqrt[3]{2\cos\frac{16\pi}9-1}$ $=\sqrt[3]{3-3\sqrt[3]9}$

Prove that:
$\frac{1}{\sqrt[3]{cos \frac{\pi}{9}}}-\frac{1}{\sqrt[3]{cos \frac{2\pi}{9}}}+\frac{1}{\sqrt[3]{cos \frac{5\pi}{9}}}=\sqrt[3]{6-6\sqrt[3]{9}}$

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a