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Difference between revisions of "1963 AHSME Problems/Problem 40"

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== Problem 40==
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== Problem ==
  
 
If <math>x</math> is a number satisfying the equation <math>\sqrt[3]{x+9}-\sqrt[3]{x-9}=3</math>, then <math>x^2</math> is between:
 
If <math>x</math> is a number satisfying the equation <math>\sqrt[3]{x+9}-\sqrt[3]{x-9}=3</math>, then <math>x^2</math> is between:

Revision as of 02:01, 7 June 2018

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

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 $\boxed{\textbf{(C)}}$.

See Also

1963 AHSC (Problems • Answer Key • Resources)
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