Find the square of:

$8x +\dfrac{3}{2}y$

Using the formula,

[∵ (x + y)² = x² + 2xy + y²]

$\Big(8x +\dfrac{3}{2}y\Big)^2\\[1em] = (8x)^2 + 2 \times 8x \times \dfrac{3}{2}y + \Big(\dfrac{3}{2}y\Big)^2\\[1em] = 64x^2 + \dfrac{48xy}{2} + \dfrac{9}{4}y^2\\[1em] = 64x^2 + 24xy + \dfrac{9}{4}y^2$

Hence, $\Big(8x +\dfrac{3}{2}y\Big)^2= 64x^2 + 24xy + \dfrac{9}{4}y^2$