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Difference between revisions of "1990 AIME Problems/Problem 2"
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== Problem ==
 
== Problem ==
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Find the value of <math>(52+6\sqrt{43})^{3/2}-(52-6\sqrt{43})^{3/2}</math>.
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== Solution 1 ==
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Suppose that <math>52+6\sqrt{43}</math> is in the form of <math>(a + b\sqrt{43})^2</math>. [[FOIL]]ing yields that <math>52 + 6\sqrt{43} = a^2 + 43b^2 + 2ab\sqrt{43}</math>. This implies that <math>a</math> and <math>b</math> equal one of <math>\pm3, \pm1</math>. The possible [[set]]s are <math>(3,1)</math> and <math>(-3,-1)</math>; the latter can be discarded since the [[square root]] must be positive. This means that <math>52 + 6\sqrt{43} = (\sqrt{43} + 3)^2</math>. Repeating this for <math>52-6\sqrt{43}</math>, the only feasible possibility is <math>(\sqrt{43} - 3)^2</math>.
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Rewriting, we get <math>(\sqrt{43} + 3)^3 - (\sqrt{43} - 3)^3</math>. Using the difference of [[cube]]s, we get that <math>[\sqrt{43} + 3\ - \sqrt{43} + 3]\ [(43 + 6\sqrt{43} + 9) + (43 - 9) + (43 - 6\sqrt{43} + 9)]</math> <math> = (6)(3 \cdot 43 + 9) = \boxed{828}</math>.
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Note: You can also just use the formula <math>(a + b)^2 = a^2 + 2ab + b^2</math> instead of [[FOIL | foiling]].
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== Solution 2 ==
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 +
The <math>3/2</math> power is quite irritating to work with so we look for a way to eliminate that. Notice that squaring the expression will accomplish that.
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Let <math>S</math> be the sum of the given expression.
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<cmath>S^2= ((52+6\sqrt{43})^{3/2}-(52-6\sqrt{43})^{3/2})^2</cmath>
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<cmath>S^2 = (52+6\sqrt{43})^{3} + (52-6\sqrt{43})^{3} - 2((52+6\sqrt{43})(52-6\sqrt{43}))^{3/2}</cmath>
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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 <math>S^2 = 685584</math> which gives <math>S=\boxed{828}</math>.
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== Solution 3 ==
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Factor as a difference of cubes.
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<cmath>\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] = </cmath>
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<cmath>\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] = </cmath>
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<cmath>\left[\left(52+6\sqrt{43}\right)^{\frac{1}{2}}-\left(52-6\sqrt{43}\right)^{\frac{1}{2}}\right]\left[104+34\right].</cmath>
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We can simplify the left factor as follows.
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<cmath>\left(52+6\sqrt{43}\right)^{\frac{1}{2}}-\left(52-6\sqrt{43}\right)^{\frac{1}{2}} = x</cmath>
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<cmath>104-2\left(52+6\sqrt{43}\right)^{\frac{1}{2}}\left(52-6\sqrt{43}\right)^{\frac{1}{2}} = x^2</cmath>
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<cmath>104-68 = x^2</cmath>
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<cmath>36 = x^2.</cmath>
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Since <math>\left(52+6\sqrt{43}\right)^{\frac{1}{2}} > \left(52-6\sqrt{43}\right)^{\frac{1}{2}}</math>, we know that <math>x=6</math>, so our final answer is <math>(6)(138) = \boxed{828}</math>.
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== Solution 4 ==
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Let <math>x=52+6\sqrt{43}</math>, <math>y=52-6\sqrt{43}</math>. Similarly to solution 2, we let
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<cmath>S=x^{\frac{3}{2}}+y^{\frac{3}{2}}</cmath>
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<cmath>\begin{align*}
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S^2&=(x^{\frac{3}{2}}+y^{\frac{3}{2}})^2\\
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&=x^3+y^3+2x^{\frac{3}{2}}y^{\frac{3}{2}}
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\end{align*}</cmath>
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The expression can be simplified as follow
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<cmath>\begin{align*}
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S^2&=x^3+y^3+2x^{\frac{3}{2}}y^{\frac{3}{2}}\\
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&=(x+y)(x^2-xy+y^2)+2(xy)^{\frac{3}{2}}\\
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&=(x+y)((x+y)^2-xy)+2\sqrt{xy}^3\\
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&=(x+y)((x+y)^2-\sqrt{xy}^2)+\sqrt{xy}^3\\
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&=(x+y)(x+y+\sqrt{xy})(x+y-\sqrt{xy})+2\sqrt{xy}^3\\
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&=104((104+34)(104-34)+2\cdot34^3\\
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&=685584
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\end{align*}</cmath>
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Thus <math>S=\sqrt{685584}=\boxed{828}</math>.
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~ 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>
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== Solution 6 ==
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(Similar to Solution 1, but expanding the cubes instead)
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Like in Solution 1, we have <math>\sqrt{52 + 6\sqrt{43}} = \sqrt{43} + 3</math> and <math>\sqrt{52 - 6\sqrt{43}} = \sqrt{43} - 3.</math>
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Therefore we have that <math>(52 + 6\sqrt{43})^{3/2} - (52 + 6\sqrt{43})^{3/2}</math> <math>= \sqrt{52 + 6\sqrt{43}}^3 - \sqrt{52 - 6\sqrt{43}}^3</math> <math>= (\sqrt{43} + 3)^3 - (\sqrt{43} - 3)^3.</math>
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From here, we use the formula <math>(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3</math> and <math>(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3</math>. Applying them to our problem we get that <math>(\sqrt{43} + 3)^3 - (\sqrt{43} - 3)^3 = (27 + 27\sqrt{43} + 9 \cdot 43 + 43\sqrt{43}) - (-27 + 27\sqrt{43} - 9*43 + 43\sqrt{43}).</math> We see that all the terms with square roots cancel, leaving us with <math>2 (27 + 9 \cdot 43) = 2 \cdot 414 = \boxed{828}.</math>
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~Yiyj1
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Note: We have that <math>\sqrt{52 - 6\sqrt{43}} = \sqrt{43} - 3</math> because we need the square root to be positive and <math>\sqrt{43} > 3</math> since <math>43</math> is obviously greater than <math>9.</math> So we have <math>\sqrt{52 - 6\sqrt{43}} = \sqrt{43} - 3.</math>
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== Video Solution ==
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https://www.youtube.com/watch?v=r96p8j0F8Fg
  
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== Solution ==
 
−
{{solution}}
 
 
== See also ==
 
== See also ==
−
* [[1990 AIME Problems]]
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{{AIME box|year=1990|num-b=1|num-a=3}}
 +
 
 +
[[Category:Intermediate Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 22:00, 26 August 2023

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 $[\sqrt{43} + 3\ - \sqrt{43} + 3]\ [(43 + 6\sqrt{43} + 9) + (43 - 9) + (43 - 6\sqrt{43} + 9)]$ $= (6)(3 \cdot 43 + 9) = \boxed{828}$. Note: You can also just use the formula $(a + b)^2 = a^2 + 2ab + b^2$ instead of foiling.

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

Solution 6

(Similar to Solution 1, but expanding the cubes instead)

Like in Solution 1, we have $\sqrt{52 + 6\sqrt{43}} = \sqrt{43} + 3$ and $\sqrt{52 - 6\sqrt{43}} = \sqrt{43} - 3.$

Therefore we have that $(52 + 6\sqrt{43})^{3/2} - (52 + 6\sqrt{43})^{3/2}$ $= \sqrt{52 + 6\sqrt{43}}^3 - \sqrt{52 - 6\sqrt{43}}^3$ $= (\sqrt{43} + 3)^3 - (\sqrt{43} - 3)^3.$

From here, we use the formula $(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$ and $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$. Applying them to our problem we get that $(\sqrt{43} + 3)^3 - (\sqrt{43} - 3)^3 = (27 + 27\sqrt{43} + 9 \cdot 43 + 43\sqrt{43}) - (-27 + 27\sqrt{43} - 9*43 + 43\sqrt{43}).$ We see that all the terms with square roots cancel, leaving us with $2 (27 + 9 \cdot 43) = 2 \cdot 414 = \boxed{828}.$

~Yiyj1

Note: We have that $\sqrt{52 - 6\sqrt{43}} = \sqrt{43} - 3$ because we need the square root to be positive and $\sqrt{43} > 3$ since $43$ is obviously greater than $9.$ So we have $\sqrt{52 - 6\sqrt{43}} = \sqrt{43} - 3.$

Video Solution

https://www.youtube.com/watch?v=r96p8j0F8Fg

See also

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