Browse · MATH

jmc

counting and probability intermediate

Problem

If

x + \frac{1}{x} = - 5

, what is

x^{3} + \frac{1}{x ^{3}}

?

Solution

To get

x^{3}

and

\frac{1}{x ^{3}}

, we cube

x + \frac{1}{x}

:

- 125 = (- 5)^{3} = (x + \frac{1}{x})^{3} = x^{3} + 3 x + \frac{3}{x} + \frac{1}{x ^{3}}

by the Binomial Theorem. Conveniently, we can evaluate

3 x + \frac{3}{x}

as

3 (x + \frac{1}{x}) = 3 (- 5) = - 15

, so

x^{3} + \frac{1}{x ^{3}} = (- 125) - (- 15) = - 110 .

Final answer

-110