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AMC 12 Mock Exam 3 Problem 5

Problem 5 of 25

A nonzero real number xx satisfies x+1x=5x+\dfrac1x=5. What is x3+1x3x^3+\dfrac1{x^3}?

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Solution

Cubing, (x+1x)3=x3+1x3+3(x+1x)\left(x+\frac1x\right)^3=x^3+\frac1{x^3}+3\left(x+\frac1x\right), so 125=x3+1x3+15125=x^3+\frac1{x^3}+15 and x3+1x3=110x^3+\frac1{x^3}=\boxed{110}. (Real solutions exist: x=5±212x=\frac{5\pm\sqrt{21}}{2}.)

Tagged: symmetric expressions in x and 1/x