Find the square of:

$3x + \dfrac{2}{y}$

Using the formula,

[∵ (x + y)² = x² + 2xy + y²]

$\Big(3x +\dfrac{2}{y}\Big)^2\\[1em] = (3x)^2 + 2 \times 3x \times \dfrac{2}{y} + \Big(\dfrac{2}{y}\Big)^2\\[1em] = 9x^2 + \dfrac{2 \times 6x}{y} + \dfrac{4}{y^2}\\[1em] = 9x^2 + \dfrac{12x}{y} + \dfrac{4}{y^2}$

Hence, $\Big(3x +\dfrac{2}{y}\Big)^2= 9x^2 + \dfrac{12x}{y} + \dfrac{4}{y^2}$