Question

Factorise: $\frac{x^{3}y}{9}-\frac{x{y}^3}{16}$

Answer 1

$\frac{x^{3}y}{9}-\frac{x{y}^3}{16}=xy(\frac{x^2}{9}-\frac{y^2}{16})=xy[{\left(\frac{x}{3}\right)}^2-{\left(\frac{y}{4}\right)}^2]$

[Using the identity $a^2-b^2=(a+b)(a-b)$]

$=xy(\frac{x}{3}-\frac{y}{4})(\frac{x}{3}+\frac{y}{4})$