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

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

@@ Line 78: / Line 78: @@
 ~Yiyj1
-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>
+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>
 == Video Solution ==

1990 AIME (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search