Question

The simplified form of the expression given below is :$\frac{\frac{y^4-x^4}{x(x+y)}-\frac{y^3}{x}}{y^2-xy+x^2}$

• A. $1$
• B. $0$
• C. $-1$
• D. $2$

Answer 1

The correct option is C $-1$

The given expression $\frac{\frac{y^4-x^4}{x(x+y)}-\frac{y^3}{x}}{y^2-xy+x^2}$ can be simplified as follows:

$\frac{\frac{y^4-x^4}{x(x+y)}-\frac{y^3}{x}}{y^2-xy+x^2}=\frac{\frac{(y^2{)}^2-(x^2{)}^2}{x(x+y)}-\frac{y^3}{x}}{y^2-xy+x^2}=\frac{\frac{(y^2-x^2)(y^2+x^2)}{x(x+y)}-\frac{y^3}{x}}{y^2-xy+x^2}(\because a^2-b^2=(a+b\left)\right(a-b\left)\right)$

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

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

Hence, $\frac{\frac{y^4-x^4}{x(x+y)}-\frac{y^3}{x}}{y^2-xy+x^2}=-1$