Find the square of:

$2m^2 -\dfrac{2}{3}n^2$

Using the formula,

[∵ (x - y)² = x² - 2xy + y²]

$\Big(2m^2 -\dfrac{2}{3}n^2\Big)^2\\[1em] = (2m^2)^2 - 2 \times 2m^2 \times \dfrac{2}{3}n^2 + \Big(\dfrac{2}{3}n^2\Big)^2\\[1em] = 4m^4 - \dfrac{2 \times 2m^2 \times 2n^2}{3} + \dfrac{4}{9}n^4\\[1em] = 4m^4 - \dfrac{8}{3}m^2n^2 + \dfrac{4}{9}n^4$

Hence, $\Big(2m^2 -\dfrac{2}{3}n^2\Big)^2= 4m^4 - \dfrac{8}{3}m^2n^2 + \dfrac{4}{9}n^4$