Question

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

• A. $1$
• B. $2xy$
• C. $xy$
• D. None of these

Answer 1

Answer: (C)

$xy$

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

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

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

$=xy$