Question

$x-\left(\frac{1}{x}\right)=3$ then the value of $x^3-\frac{1}{x^3}$ is?

Answer 1

Find the value of $x^3-\frac{1}{x^3}$ using algebraic identities

Consider,

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

Cubing both sides, we get

$$\begin{gathered} {\left(x-\frac{1}{x}\right)}^3=3^3 \\ =27 \end{gathered}$$

Now,

$$\begin{gathered} {\left(x-\frac{1}{x}\right)}^3=x^3-\frac{1}{x^3}-3\times x\times \frac{1}{x}\left(x-\frac{1}{x}\right)\left[\because {\left(a-b\right)}^3=a^3-b^3-3ab\left(a-b\right)\right] \\ \Rightarrow 27=x^3-\frac{1}{x^3}-3\times 1\times 3\left[\because x-\frac{1}{x}=3\right] \\ \Rightarrow 27=x^3-\frac{1}{x^3}-9 \\ \Rightarrow x^3-\frac{1}{x^3}=27+9 \\ \therefore x^3-\frac{1}{x^3}=36 \end{gathered}$$

Thus, the value of $x^3-\frac{1}{x^3}=36$ .