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Difference between revisions of "1990 AIME Problems/Problem 2"

(→‎Solution 1)
(Added Solution 5)
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~ Nafer
 
~ Nafer
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== Solution 5 ==
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(Similar to Solution 3, but with substitution)
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Let <math>a=\sqrt{52+6\sqrt{43}}</math> and <math>b=\sqrt{52-6\sqrt{43}}.</math> We want to find <math>a^3-b^3=(a-b)(a^2+ab+b^2).</math>
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We have
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<cmath>a^2+b^2=102,\text{ and}</cmath>
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<cmath>ab=\sqrt{(52+6\sqrt{43})(52-6\sqrt{43})}=\sqrt{1156}=34.</cmath>
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Then, <math>(a-b)^2=a^2+b^2-2ab=104-2\cdot 34= 36\implies a-b=6.</math>
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Our answer is
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<cmath>a^3-b^3=(a-b)(a^2+b^2+ab)=6\cdot 138=\boxed{828.}</cmath>
  
 
== See also ==
 
== See also ==

Revision as of 10:16, 7 October 2021

Problem

Find the value of $(52+6\sqrt{43})^{3/2}-(52-6\sqrt{43})^{3/2}$.

Solution 1

Suppose that $52+6\sqrt{43}$ is in the form of $(a + b\sqrt{43})^2$. FOILing yields that $52 + 6\sqrt{43} = a^2 + 43b^2 + 2ab\sqrt{43}$. This implies that $a$ and $b$ equal one of $\pm3, \pm1$. The possible sets are $(3,1)$ and $(-3,-1)$; the latter can be discarded since the square root must be positive. This means that $52 + 6\sqrt{43} = (\sqrt{43} + 3)^2$. Repeating this for $52-6\sqrt{43}$, the only feasible possibility is $(\sqrt{43} - 3)^2$.

Rewriting, we get $(\sqrt{43} + 3)^3 - (\sqrt{43} - 3)^3$. Using the difference of cubes, we get that $[\sqrt{43} + 3\ - \sqrt{43} + 3]\ [(43 + 6\sqrt{43} + 9) + (43 - 9) + (43 - 6\sqrt{43} + 9)]$ $= (6)(3 \cdot 43 + 9) = \boxed{828}$.

Solution 2

The $3/2$ power is quite irritating to work with so we look for a way to eliminate that. Notice that squaring the expression will accomplish that. Let $S$ be the sum of the given expression. \[S^2= ((52+6\sqrt{43})^{3/2}-(52-6\sqrt{43})^{3/2})^2\] \[S^2 = (52+6\sqrt{43})^{3} + (52-6\sqrt{43})^{3} - 2((52+6\sqrt{43})(52-6\sqrt{43}))^{3/2}\] After doing the arithmetic (note that the first two terms will have some cancellation and that the last term will simplify quickly using difference of squares), we arrive at $S^2 = 685584$ which gives $S=\boxed{828}$.

Solution 3

Factor as a difference of cubes. \[\left[\left(52+6\sqrt{43}\right)^{\frac{1}{2}}-\left(52-6\sqrt{43}\right)^{\frac{1}{2}}\right]\left[\left(\left(\left(52+6\sqrt{43}\right)^{\frac{1}{2}}\right)^2+\left(52+6\sqrt{43}\right)^{\frac{1}{2}}\left(52-6\sqrt{43}\right)^{\frac{1}{2}}+\left(\left(52-6\sqrt{43}\right)^{\frac{1}{2}}\right)^2\right)\right] =\] \[\left[\left(52+6\sqrt{43}\right)^{\frac{1}{2}}-\left(52-6\sqrt{43}\right)^{\frac{1}{2}}\right]\left[104+\left(52^2-\left(36\right)\left(43\right)\right)^{\frac{1}{2}}\right] =\] \[\left[\left(52+6\sqrt{43}\right)^{\frac{1}{2}}-\left(52-6\sqrt{43}\right)^{\frac{1}{2}}\right]\left[104+34\right].\] We can simplify the left factor as follows. \[\left(52+6\sqrt{43}\right)^{\frac{1}{2}}-\left(52-6\sqrt{43}\right)^{\frac{1}{2}} = x\] \[104-2\left(52+6\sqrt{43}\right)^{\frac{1}{2}}\left(52-6\sqrt{43}\right)^{\frac{1}{2}} = x^2\] \[104-68 = x^2\] \[36 = x^2.\] Since $\left(52+6\sqrt{43}\right)^{\frac{1}{2}} > \left(52-6\sqrt{43}\right)^{\frac{1}{2}}$, we know that $x=6$, so our final answer is $(6)(138) = \boxed{828}$.


Solution 4

Let $x=52+6\sqrt{43}$, $y=52-6\sqrt{43}$. Similarly to solution 2, we let \[S=x^{\frac{3}{2}}+y^{\frac{3}{2}}\] \begin{align*} S^2&=(x^{\frac{3}{2}}+y^{\frac{3}{2}})^2\\ &=x^3+y^3+2x^{\frac{3}{2}}y^{\frac{3}{2}} \end{align*} The expression can be simplified as follow \begin{align*} S^2&=x^3+y^3+2x^{\frac{3}{2}}y^{\frac{3}{2}}\\ &=(x+y)(x^2-xy+y^2)+2(xy)^{\frac{3}{2}}\\ &=(x+y)((x+y)^2-xy)+2\sqrt{xy}^3\\ &=(x+y)((x+y)^2-\sqrt{xy}^2)+\sqrt{xy}^3\\ &=(x+y)(x+y+\sqrt{xy})(x+y-\sqrt{xy})+2\sqrt{xy}^3\\ &=104((104+34)(104-34)+2\cdot34^3\\ &=685584 \end{align*} Thus $S=\sqrt{685584}=\boxed{828}$.

~ Nafer

Solution 5

(Similar to Solution 3, but with substitution)

Let $a=\sqrt{52+6\sqrt{43}}$ and $b=\sqrt{52-6\sqrt{43}}.$ We want to find $a^3-b^3=(a-b)(a^2+ab+b^2).$

We have \[a^2+b^2=102,\text{ and}\] \[ab=\sqrt{(52+6\sqrt{43})(52-6\sqrt{43})}=\sqrt{1156}=34.\] Then, $(a-b)^2=a^2+b^2-2ab=104-2\cdot 34= 36\implies a-b=6.$

Our answer is \[a^3-b^3=(a-b)(a^2+b^2+ab)=6\cdot 138=\boxed{828.}\]

See also

1990 AIME (Problems • Answer Key • Resources)
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Problem 3
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