Question

Find the cube of $(x-\frac{1}{x})$

Answer 1

Using the identity $(a-b{)}^3=a^3-b^3-3a^{2}b+3a{b}^2$

Therefore,

${(x-\frac{1}{x})}^3=(x{)}^3-{\left(\frac{1}{x}\right)}^3-(3\times (x{)}^2\times \frac{1}{x})+(3\times x\times {\left(\frac{1}{x}\right)}^2)=x^3-\frac{1}{x^3}-3x+\frac{3}{x}$

Hence, ${(x-\frac{1}{x})}^3=x^3-\frac{1}{x^3}-3x+\frac{3}{x}$.