Question

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

$55$ and $65$

$65$ and $75$

$75$ and $85$

$85$ and $95$

$95$ and $105$

Answer: C

Concepts:radical sum and difference of cubes symmetry (algebra)

Difficulty rating: 2290

Small Hint:

Set $u=\sqrt[3]{x+9}$ and $v=\sqrt[3]{x-9}$

Big Hint:

Use both $u-v=3$ and $u^3-v^3=18$ to find $uv$

Solution:

Let $u=\sqrt[3]{x+9}$ and $v=\sqrt[3]{x-9}.$ Then $u-v=3$ and $\begin{aligned} 18&=u^3-v^3\\ &=(u-v)(u^2+uv+v^2), \end{aligned}$ so $u^2+uv+v^2=6.$ Comparing with $(u-v)^2=9$ gives $uv=-1.$ Hence $(u+v)^2=5.$ Cubing $2u=3\pm\sqrt5$ gives $x=\pm4\sqrt5,$ and either way $x^2=80,$ between $75$ and $85.$

Therefore, the correct answer is C.

1963 AMC 12 Problem 40

Answer: C

Difficulty rating: 2290

Small Hint:

Big Hint:

Solution:

Problem 40 in Other Years