Question

The simplified value of [1-(2xy/x2+y2)] / [(x3+y3/x-y)3xy] is

Answer 1

Answer:

After simplification we get, $\frac{(x-y)^3}{3x^6y+3x^3y^4+3x^4y^3+3xy^6}$

Step-by-step explanation:

Given Expression:

$\frac{1-\frac{2xy}{x^2+y^2}}{(\frac{x^3+y^3}{x-y})3xy}$

We need to simplify the given expression

Consider,

$\frac{1-\frac{2xy}{x^2+y^2}}{(\frac{x^3+y^3}{x-y})3xy}$

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

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

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

$=\frac{(x-y)^2}{x^2+y^2}\times\frac{x-y}{3x^4y+3xy^4}$

$=\frac{(x-y)^3}{x^2(3x^4y+3xy^4)+y^2(3x^4y+3xy^4)}$

$=\frac{(x-y)^3}{3x^6y+3x^3y^4+3x^4y^3+3xy^6}$

Therefore, after simplification we get, $\frac{(x-y)^3}{3x^6y+3x^3y^4+3x^4y^3+3xy^6}$