Find the cube of:

$2x +\dfrac{1}{x}$

$2x + \dfrac{1}{x}$

Using the formula,

(x + y)³ = x³ + y³ + 3x²y + 3xy²

$= (2x)^3 + \Big(\dfrac{1}{x}\Big)^3 + 3 \times (2x)^2 \times \Big(\dfrac{1}{x}\Big) + 3 \times 2x \times \Big(\dfrac{1}{x}\Big)^2\\[1em] = 8x^3 + \Big(\dfrac{1}{x^3}\Big) + \Big(\dfrac{12x^2}{x}\Big) + \Big(\dfrac{6x}{x^2}\Big)\\[1em] = 8x^3 + \Big(\dfrac{1}{x^3}\Big) + 12x + \dfrac{6}{x}\\[1em]$

Hence, $2x + \dfrac{1}{x} = 8x^3 + \Big(\dfrac{1}{x^3}\Big) + 12x + \dfrac{6}{x}$