Question

Represent the following mixed infinite decimal periodic fractions as common fractions:
$\frac{x^3+y^3}{x+y}:(x^2-y^2)+\frac{2y}{x+y}-\frac{xy}{x^2-y^2}$

Answer 1

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

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

$=x^2+y^2-xy$

and $x^2-y^2+\frac{2y}{x+y}-\frac{xy}{x^2-y^2}$

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

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

$=\frac{x^4+y^4-2x^2{y}^2+x{y}^2-x^{2}y+2y}{x^2-y^2}$

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

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

$=\frac{x^4-y^4-x^{3}y+x{y}^3}{x^4-y^4-2x^2{y}^2+x{y}^2-x^{2}y+2y}.$

Hence, the answer is $\frac{x^4-y^4-x^{3}y+x{y}^3}{x^4-y^4-2x^2{y}^2+x{y}^2-x^{2}y+2y}.$