Question

Find the cube of
$\left(v\right)(2x+\frac{1}{x})$

Answer 1

Applying the formula of cube of sum of two terms:
${(2x+\frac{1}{x})}^3=(2x{)}^3+3\times (2x{)}^2\times \left(\frac{1}{x}\right)+3\times 2x\times {\left(\frac{1}{x}\right)}^2+{\left(\frac{1}{x}\right)}^3$
$[\because (a+b{)}^3=a^3+b^3+3\times a^2\times b+3\times a\times b^2]$
$=8x^3+12x+\left(\frac{6}{x}\right)+\frac{1}{x^3}$
Hence, the cube of $(2x+\frac{1}{x})$ is $8x^3+12x+\left(\frac{6}{x}\right)+\frac{1}{x^3}$.