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1963 AHSME Problems/Problem 40

Revision as of 10:09, 6 January 2021 by Pfalcon (talk | contribs)

Problem

If $x$ is a number satisfying the equation $\sqrt[3]{x+9}-\sqrt[3]{x-9}=3$, then $x^2$ is between:

$\textbf{(A)}\ 55\text{ and }65\qquad \textbf{(B)}\ 65\text{ and }75\qquad \textbf{(C)}\ 75\text{ and }85\qquad \textbf{(D)}\ 85\text{ and }95\qquad \textbf{(E)}\ 95\text{ and }105$

Solution 1

Let $a = \sqrt[3]{x + 9}$ and $b = \sqrt[3]{x - 9}$. Cubing these equations, we get $a^3 = x + 9$ and $b^3 = x - 9$, so $a^3 - b^3 = 18$. The left-hand side factors as \[(a - b)(a^2 + ab + b^2) = 18.\]

However, from the given equation $\sqrt[3]{x + 9} - \sqrt[3]{x - 9} = 3$, we get $a - b = 3$. Then $3(a^2 + ab + b^2) = 18$, so $a^2 + ab + b^2 = 18/3 = 6$.

Squaring the equation $a - b = 3$, we get $a^2 - 2ab + b^2 = 9$. Subtracting this equation from the equation $a^2 + ab + b^2 = 6$, we get $3ab = -3$, so $ab = -1$. But $a = \sqrt[3]{x + 9}$ and $b = \sqrt[3]{x - 9}$, so $ab = \sqrt[3]{(x + 9)(x - 9)} = \sqrt[3]{x^2 - 81}$, so $\sqrt[3]{x^2 - 81} = -1$. Cubing both sides, we get $x^2 - 81 = -1$, so $x^2 = 80$. The answer is $\boxed{\textbf{(C)}}$.


Solution 2

$\sqrt[3]{x+9}-\sqrt[3]{x-9}-3=0$ i.e, $\sqrt[3]{x+9}+\sqrt[3]{-x+9}+(-3)=0$ 

if the sum of three numbers is zero, then their sum of cubes is thrice the product of each number.
then, $x+9-x+9+27=3(\sqrt[3]{x+9})(\sqrt[3]{9-x})(-3)$ . by solving this, we get $-9=-9\sqrt[3]{9^2-x^2}$
$x^2=80$. this gives the step what we had done in solution 1.The answer is $\boxed{\textbf{(C)}}$.

Solution 3

Cubing both sides to get rid of the messy cube roots, we get \[x+9-3\sqrt[3]{(x+9)^2(x-9)}+3\sqrt[3]{(x+9)(x-9)^2}-x+9=27,\]so \[18-3\sqrt[3]{(x-9)(x+9)}(\sqrt[3]{x+9}-\sqrt[3]{x-9})=27\implies 3\sqrt[3]{x^2-81}=-3.\] Dividing both sides by 3 and cubing, we find $x^2=80$, which is between $\boxed{(C)75\text{ and }85}$.


See Also

1963 AHSC (Problems • Answer Key • Resources)
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