1963 AHSME Problems/Problem 40

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 $x^2 = \boxed{80}$ . The answer is (C).

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1963 AHSME Problems/Problem 40