Question

If $x-\frac{1}{x}=3$, then the value $x^3-\frac{1}{x^3}$ is

• A. $27$
• B. $32$
• C. $36$
• D. $42$

Answer 1

The correct option is C $36$

Since ${\left(a-\frac{1}{a}\right)}^3=a^3-\frac{1}{a^3}-3\left(a-\frac{1}{a}\right)$, therefore, consider ${\left(x-\frac{1}{x}\right)}^3$ and solve:

${\left(x-\frac{1}{x}\right)}^3=x^3-\frac{1}{x^3}-3\left(x-\frac{1}{x}\right)\Rightarrow {\left(x-\frac{1}{x}\right)}^3=(x^3-\frac{1}{x^3})-3\left(x-\frac{1}{x}\right)\Rightarrow (x^3-\frac{1}{x^3})={\left(x-\frac{1}{x}\right)}^3+3\left(x-\frac{1}{x}\right)$

It is given that $x-\frac{1}{x}=3$, therefore,

$x^3-\frac{1}{x^3}={\left(3\right)}^3+3\left(3\right)=27+9=36$

Hence, $x^3-\frac{1}{x^3}=36$.