Question

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

Answer 1

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

Therefore,

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

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