Question

Simplify the expression :

$\left[\frac{x^3+y^3}{{(x-y)}^2+3xy}\right]÷\left[\frac{{(x+y)}^2-3xy}{x3-y^3}\right]\times \frac{xy}{x^2-y^2}$

• A. $xy$
• B. $x^{2}y$
• C. $x{y}^2$
• D. $x^2{y}^2$

Answer 1

The correct option is A $xy$

$\left(\frac{x^3+y^3}{{(x-y)}^2+3xy}\right)÷\left(\frac{{(x+y)}^2-3xy}{x^3-y^3}\right)\times \frac{xy}{x^2-y^2}$

$=\left(\frac{(x+y)(x^2-xy+y^2)}{x^2-2xy+y^{2}3xy}\right)÷\left(\frac{x^2+y^2+2xy-3xy}{(x-y)(x^2+xy+y^2)}\right)\times \frac{xy}{x^2-y^2}$

$=\frac{(x+y)(x^2-xy+y^2)}{(x^2+xy+y^2)}\times \frac{(x-y)(x^2+xy+y^2)}{(x^2-xy+y^2)}\times \frac{xy}{(x+y)(x-y)}$

$=xy$