Sum of First, Second, and Third Powers

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Brut3Forc3

1948 posts

Brut3Forc3

#1 h Mar 7, 2010, 12:20 AM • 8 Y

Y by jhu08, ImSh95, ryanbear, Amir Hossein, Rounak_iitr, Adventure10, cubres, and 1 other user

Determine all roots, real or complex, of the system of simultaneous equations
$\begin{align*} x+y+z &= 3, \\ x^2+y^2+z^2 &= 3, \\ x^3+y^3+z^3 &= 3.\end{align*}$

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modularmarc101

2208 posts

modularmarc101

#2 h Mar 7, 2010, 1:03 AM • 9 Y

Y by smo, Killer_Instinct, jhu08, ImSh95, Amir Hossein, Adventure10, kiyoras_2001, cubres, and 1 other user

We begin by manipulating the given equations to find other useful relations between the variables.

$(\sum x)^2 = \sum x^2 + 2 \sum xy$

$3^2 = 3 + 2 \sum xy$

$\sum xy = 3$

Now using the following identity,

$\sum x^3 - 3xyz = (\sum x)(\sum x^2 - \sum xy)$

$3 - 3xyz = 3(0)$

$xyz = 1$

Now, let $x,y,$ and $z$ be the roots of a cubic polynomial $P(t)$ . Then, by Vieta's Formulas, we have that

$P(t) = t^3 - 3t^2 + 3t - 1$

$P(t) = (t - 1)^3$

Therefore, $\boxed{x = y = z = 1}$ is the only solution.

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daydreamer1116

116 posts

daydreamer1116

#3 h Mar 13, 2010, 4:35 AM • 5 Y

Y by ImSh95, Adventure10, Mango247, cubres, and 1 other user

If we square the $x + y + z = 3$ , then we get
$9 = x^2 + y^2 + z^2 + 2(xy + yz + xz)$
$9 = 3 + 2(xy + yz + xz)$
$xy + yz + xz = 3$

If we multiply this by $x + y + z$ , then we have
$9 = (x^2y + xy^2 + y^2 + yz^2 + x^2z + xz^2) + 3xyz$ and we substitute the contents in the parenthesis with $a$ for convenience, then we have
$9 = a + 3xyz$
$xyz = \frac {9 - a}{3}$

If we multiply $x^2 + y^2 + z^2$ by $x + y + z$ , then we get
$x^3 + y^3 + z^3 + (x^2y + xy^2 + y^2 + yz^2 + x^2z + xz^2) = 9$
$x^3 + y^3 + z^3 + a = 9$
$3 + a = 9$
$a = 6$

If we plug this back in to get xyz, we have
$xyz = 1$

Notice that we now have $x + y + z$ , $xy+yz+xz$ , and $xyz$ and if we use $x$ , $y$ , and $z$ as roots of a cubic equation, then we have
$a^3 + 3a^2 + 3a + 1 = 0$
$(a - 1)^3 = (a - x)(a - y)(a - z)$
Therefore $x = y = z = 1$

and I guess that's the only solution... correct me if I'm wrong please... oh yeah, modularmarc101 got the same thing so I guess I can't be that off...

This post has been edited 2 times. Last edited by daydreamer1116, Mar 13, 2010, 4:40 AM

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Brut3Forc3

1948 posts

Brut3Forc3

#4 h Mar 13, 2010, 4:38 AM • 5 Y

Y by ImSh95, Adventure10, Mango247, cubres, and 1 other user

daydreamer1116 wrote:

Notice that we now have $x + y + z$ , $x^2y + xy^2 + y^2 + yz^2 + x^2z + xz^2$ , and $xyz$ and if we use $x$ , $y$ , and $z$ as roots of a cubic equation, then we have...

I think you meant $xy+yz+zx$ for the middle one. If so, then the solution is correct.

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daydreamer1116

116 posts

daydreamer1116

#5 h Mar 13, 2010, 4:41 AM • 5 Y

Y by ImSh95, Adventure10, Mango247, cubres, and 1 other user

Thanks. I think I'm good now

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professordad

4549 posts

professordad

#6 h Dec 31, 2011, 2:34 AM • 5 Y

Y by ImSh95, Adventure10, Mango247, cubres, and 1 other user

Sorry for reviving; what is wrong with this solution?

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fishythefish

4001 posts

fishythefish

#7 h Dec 31, 2011, 5:22 AM • 4 Y

Y by ImSh95, Adventure10, cubres, and 1 other user

Well, everyone else got $x=y=z=1$ , and I agree, so I'm not sure what you mean. Computer confirms that that is the only solution.

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tonypr

762 posts

tonypr

#8 h Dec 31, 2011, 5:38 AM • 6 Y

Y by professordad, ImSh95, Adventure10, ehuseyinyigit, cubres, and 1 other user

professordad wrote:

We have $\left(\sum x\right)^2 = \left(\sum x^2\right) + 2\left(\sum xy\right)$ , so $xy + yz + xz = 3$ .

Now $x^2 + y^2 + z^2 - xy - yz - xz = 0$ , so $(x - y)^2 + (y - z)^2 + (x - z)^2 = 0$ . All squares must equal 0, so $x = y = z = 1$ .

When $x,y$ and $z$ are complex numbers, we don't immediately have
$(x-y)^2\ge 0$
(and cyclic permutations)

So it's not immediate that
$(x - y)^2 + (y - z)^2 + (x - z)^2 = 0\implies x=y=z$

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Virgil Nicula

7054 posts

Virgil Nicula

#9 h Dec 31, 2011, 1:25 PM • 5 Y

Y by ImSh95, teomihai, Adventure10, cubres, and 1 other user

You can use the Chebyshev's ineguality and $\underline{(x+y+z)(x^2+y^2+z^2)}=3\cdot 3=\underline{3\cdot (x^3+y^3+z^3)}\iff x=y=z$ .

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mavropnevma

15142 posts

mavropnevma

#10 h Dec 31, 2011, 2:25 PM • 5 Y

Y by ImSh95, Adventure10, Mango247, Jack_w, cubres

You cannot use Chebyshev's inequality (which makes no sense) when the variables might be complex numbers.

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Virgil Nicula

7054 posts

Virgil Nicula

#11 h Dec 31, 2011, 2:47 PM • 3 Y

Y by ImSh95, Adventure10, cubres

OK. I don't see complex numbers, sorry ... Thank you. Happy new year !

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proshi

2187 posts

proshi

#12 h Apr 25, 2018, 10:19 PM • 4 Y

Y by ImSh95, Adventure10, Mango247, cubres

Let $P(x)=t^3+at^2+bt+c=0$ be a cubic polynomial with roots $x,y,z.$ By Vieta's we know that $a=-1(x+y+z) \implies a=-3.$ We also know that $b=xy+xz+yz.$ We can now use Newton's sums to derive $b.$ Newton's sums tells us that $(x+y+z)(x+y+z)-2(xy+xz+yz)=x^2+y^2+z^2.$ We can plug in the values of $x+y+z$ and $x^2+y^2+z^2$ to get $3\cdot 3-2(xy+xz+yz)=3 \implies xy+xz+yz=3.$ Hence, $b=3.$

We can use Vieta's and Newton's sums again to derive $c$ as we know that $c=-1(xyz).$ So, by Newton's sums, it follows that $(x+y+z)(x^2+y^2+z^2)-(xy+xz+yz)(x+y+z)+3(xyz)=x^3+y^3+z^3.$ We know what $(x+y+z), (x^2+y^2+z^2), (xy+xz+yz)$ and $(x^3+y^3+z^3)$ are so we plug them in

$3\cdot 3-3\cdot 3+3(xyz)=3 \implies xyz=1.$ Hence, $c=-1.$

Now our cubic polynomial is $t^3-3t^2+3t-1=0.$ That factors into $(t-1)^3=0.$ So, $t=1$ . Therefore, our only roots to the system of simultaneous equations are $\boxed{x=y=z=1}.$

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OlympusHero

17020 posts

OlympusHero

#13 h Apr 12, 2021, 2:32 PM • 2 Y

Y by ImSh95, cubres

Solution

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Louis_Vuitton

223 posts

Louis_Vuitton

#14 h Apr 12, 2021, 2:36 PM • 2 Y

Y by ImSh95, cubres

can you use the newton sum thing?

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franzliszt

23531 posts

franzliszt

#15 h Jun 29, 2021, 6:30 PM • 3 Y

Y by ImSh95, Jack_w, cubres

Louis_Vuitton wrote:

can you use the newton sum thing?

Yes!

The answer is $(x,y,z)=(1,1,1)$ which clearly works. Now we will show that it is the only solution.

Let $x,y,z$ be the roots of some cubic polynomial $x^3+ax^2+bx+c$ . Define the sums $P_k=x^k+y^k+z^k$ . Then by Newton Sums, $\begin{align*}P_1+a&=0\\P_2+aP_1+2b&=0\\ P_3+aP_2+bP_1+3c&=0\end{align*}$ or $\begin{align*}a&=-P_1\\b&=-\frac{P_2+aP_1}2\\ c&=-\frac{P_3+aP_2+bP_1}3.\end{align*}$
We are already given $P_1=P_2=P_3=3$ , so we can find $a=-3$ which tells us that $b=3$ which tells us that $c=-1$ .

Hence, $x,y,z$ are the solutions to $f(t)=t^3-3t^2+3t-1.$ Factoring, $f(t)=t^3-3t^2+3t-1=(t-1)^3$ so $f(x)$ has a triple root at $t=1$ . Thus, we find that $(x,y,z)=(1,1,1)$ is the only solution to the original system.

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Significant

211 posts

Significant

#35 h Aug 5, 2023, 7:36 PM • 1 Y

Y by cubres

Denote $a, b,$ and $c$ as follows,
$\begin{align*} a & = x+y+z, \\ b & = xy+yz+zx, \\ c & = xyz. \end{align*}$ Our given equations become
$\begin{align*} 3 & = a, \\ 3 & = a^2-2b, \\ 3 & = a(a^2-b)+3c =a^3-ab+3c. \end{align*}$ Solving for $b$ and $c$ , we obtain $b=3$ and $c=1$ .
Now let $P(x)$ be the cubic polynomial with roots $x, y,$ and $z$ . By Vieta's,
$\begin{align*} P(x) & =x^3-3x^2+3x-1, \\ & =(x-1)^3. \end{align*}$ Therefore, $\boxed{x=y=z=1}$ is our only solution.

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AlanLG

241 posts

AlanLG

#36 h Aug 8, 2023, 5:52 PM • 1 Y

Y by cubres

Let $P(t)=t^3+a_2t^2+a_1t+a_0$ the monic pollynomial with roots $x,y,z$ ; by Newton Sums
$\begin{align} 0 &= 1 \cdot (x+y+z)+a_2\Rightarrow a_2=-3\nonumber \\ &=1\cdot (x^2+y^2+z^2)+a_2(x+y+z)+a_1\cdot 2\Rightarrow a_1=3 \nonumber \\ &=1\cdot(x^3+y^3+z^3)+a_2(x^2+y^2+z^2)+a_1(x+y+z)+3a_0\Rightarrow a_0=-1\nonumber \end{align}$ Therefore $P(t)=t^3-3t^2+3t-1=(t-1)^3\Rightarrow x=y=z=1$

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mathmax12

6001 posts

mathmax12

#37 h Sep 4, 2023, 9:31 PM • 1 Y

Y by cubres

Let, $q=x+y+z, r=xy+xz+yz, s=xyz,$ then $3=q, 3=q^2-2r, 3=q^3-qr+3s.$ Note, that $r=\frac{q^2-3}{2}=\frac{9-3}{2}=3.$ And, we find that $s=1.$ Now, by Vieta's this is the cubic polynomial, $x^3-3x^2+3x-1=(x-1)^3$ , so we must have $x=y=z=1.$

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joshualiu315

2513 posts

joshualiu315

#38 h Oct 13, 2023, 4:48 PM • 1 Y

Y by cubres

We will use standard SQP notation:

$\begin{align*} s &= x+y+z \\ q &= xy+yz+zx \\ p &= xyz \end{align*}$
From the first equation, we immediately know $s=3$ . Then, we have $x^2+y^2+z^2=s^2-2q$ , so we can solve that $q=3$ . To apply the last equation, notice that

$x^3+y^3+z^3=3xyz+(x+y+z)(x^2+y^2+z^2-xy-yz-xz)$
Simplifying this and solving, we get $p=1$ .

By Vieta's, we see that $x$ , $y$ , and $z$ are the roots of polynomial $P(t)=t^3-3t^2+3t-1=(t-1)^3$ . Thus, the only triple $(x,y,z)$ is $\boxed{(1,1,1)}$ .

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Cusofay

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Cusofay

#39 h Dec 1, 2023, 11:08 AM • 1 Y

Y by cubres

By using Newton's sum, we can find that $x$ , $y$ and $z$ are roots of the polynomial $P(t)=(t-1)^3$ . This implies that $x=y=z=1$ is the only solution to this system of equations.

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peppapig_

280 posts

peppapig_

#40 h Dec 12, 2023, 1:03 AM • 1 Y

Y by cubres

I claim that the only possible solution is $(1,1,1)$ .

Note that from the second equation, we get that
$xy+yz+zx=\frac{(x+y+z)^2-(x^2+y^2+z^2)}{2}=\frac{9-3}{2}=3,$ and using this, along with the third equation, we get that
$3-3xyz=(x^3+y^3+z^3)-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)=3(0)=0 \iff xyz=1,$ which gives us the system of equations
$x+y+z=3,$ $xy+yz+zx=3,$ $xyz=1,$ which implies that $x$ , $y$ , $z$ should all be roots of $f(a)=a^3-3a^2+3a-1=(a-1)^3$ by Vieta's. This gives us that the only solution to $(x,y,z)$ is $(1,1,1)$ , finishing the problem.

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David_Kim_0202

384 posts

David_Kim_0202

#41 h Dec 12, 2023, 1:36 PM • 1 Y

Y by cubres

since, $x+y+z=3, x²+y²+z²=3$ , we can say $xy+yz+zx = 3$ .
then, by subtracting $3xyz$ by both sides by the third equation,
$x³+y³+z³-3xyz=(x+y+z)(x²+y²+z²-xy-yz-zx)=0$ also, $x³+y³+z³=3$ so $xyz=1$ .
therefore we know all of the sum, pair sum, product of $x, y, x$ , we can know these 3 variables are the roots of a polynomial $t³-3t²+3t-1=0$
therefore the answer is $\boxed{x=y=z=1}$

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blueberryfaygo_55

340 posts

blueberryfaygo_55

#42 h Mar 26, 2024, 4:15 AM • 1 Y

Y by cubres

From equation $(1)$ and equation $(2)$ , we obtain $xy+yz+zx=3$ . It follows that $x^2+y^2+z^2=xy+yz+zx$ , and multiplying both sides by $2$ and rearranging to factor into binomials yields $(x-y)^2+(y-z)^2+(z-x)^2 = 0$ so we must have $x=y=z=1$ as the only solution.

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AshAuktober

990 posts

AshAuktober

#43 h Mar 27, 2024, 4:15 AM • 1 Y

Y by cubres

No way is this USAMO.
From standard identities for $(a+b+c)^2$ and $a^3 + b^3 + c^3 -3abc$ , we can obtain $xy + yz + zx = 3, xyz = 1,$ so $x, y, z$ are the zeroes of $t^3 - 3t^2 + 3t - 1 = (t-1)^3,$ and we get $x = y = z = 1.$ Since it works, we're done. $\square$

This post has been edited 1 time. Last edited by AshAuktober, Aug 10, 2024, 4:20 AM

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AshAuktober

990 posts

AshAuktober

#44 h Mar 27, 2024, 4:16 AM • 1 Y

Y by cubres

blueberryfaygo_55 wrote:

From equation $(1)$ and equation $(2)$ , we obtain $xy+yz+zx=3$ . It follows that $x^2+y^2+z^2=xy+yz+zx$ , and multiplying both sides by $2$ and rearranging to factor into binomials yields $(x-y)^2+(y-z)^2+(z-x)^2 = 0$ so we must have $x=y=z=1$ as the only solution.

How do you get the last step when $x, y, z$ need not be real?

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Mr.Sharkman

496 posts

Mr.Sharkman

#45 h Apr 16, 2024, 10:32 PM • 1 Y

Y by cubres

Solution

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Markas

105 posts

Markas

#46 h May 5, 2024, 1:26 PM • 1 Y

Y by cubres

Lets assume that x, y, z are the roots of a cubic polynomial $ax^3 + bx^2 + cx + d$ . By the condition we have $p_1 = 3$ , $p_2 = 3$ , $p_3 = 3$ $\Rightarrow$ by Newton's sums we have that 0 = 3a + b = 3a + 3b + 2c = 3a + 3b + 3c + 3d. Also we can assume a = 1 $\Rightarrow$ from this we get b = -3, c = 3, d = -1 $\Rightarrow$ the polynomial which roots are x, y and z is $x^3 - 3x^2 + 3x - 1$ . Also $x^3 - 3x^2 + 3x - 1 = (x - 1)^3$ $\Rightarrow$

x = y = z = 1.

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eg4334

631 posts

eg4334

#47 h Aug 7, 2024, 11:55 PM • 1 Y

Y by cubres

Consider the polynomial $(t-x)(t-y)(t-z) = 0$ . The $t^2$ coefficient is $-3$ here by vietas and the $t$ coefficient is $\frac{3^2-3}{2} = 3$ by vietas as well. Also $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)$ so we can find that $xyz=1$ . Thus the polynomial is $x^3-3x^2+3x-1 = (x-1)^3$ so $\boxed{x=y=z=1}$ is the only solution.

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megahertz13

3182 posts

megahertz13

#48 h Nov 4, 2024, 4:35 PM • 1 Y

Y by cubres

The answer is $(1,1,1)$ only. Let $a=x+y+z$ $b=xy+yz+zx$ $c=xyz.$ Note that $a=3$ , $a^2-2b=3$ , and $3-3c=a(a^2-3b)\implies (a,b,c)=(3,3,1).$ By Vieta's, we find that $x$ , $y$ , and $z$ are the roots of the cubic $a^3-3a^2+3a-1=(a-1)^3=0,$ so $(x,y,z)=(1,1,1)$ is the only solution.

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cubres

114 posts

cubres

#49 h Apr 12, 2025, 3:20 PM

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Solution

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