Solve (b+a)(2b-a)(a-b)^2/(b-a)(b^2-a^2)(2b-a)-1

$$\frac { ( b + a ) ( 2 b - a ) ( a - b ) ^ { 2 } } { ( b - a ) ( b ^ { 2 } - a ^ { 2 } ) ( 2 b - a ) } - 1$$

Evaluate

$$0$$

Short Solution Steps

Cancel out -a+2b in both numerator and denominator.

$$\frac{\left(a+b\right)\left(a-b\right)^{2}}{\left(-a+b\right)\left(-a^{2}+b^{2}\right)}-1$$

Factor the expressions that are not already factored in \frac{\left(a+b\right)\left(a-b\right)^{2}}{\left(-a+b\right)\left(-a^{2}+b^{2}\right)}.

$$\frac{\left(a+b\right)\left(a-b\right)^{2}}{\left(a-b\right)\left(-a+b\right)\left(-a-b\right)}-1$$

Extract the negative sign in a+b.

$$\frac{-\left(-a-b\right)\left(a-b\right)^{2}}{\left(a-b\right)\left(-a+b\right)\left(-a-b\right)}-1$$

Cancel out \left(a-b\right)\left(-a-b\right) in both numerator and denominator.

$$\frac{-\left(a-b\right)}{-a+b}-1$$

Extract the negative sign in a-b.

$$\frac{-\left(-1\right)\left(-a+b\right)}{-a+b}-1$$

Cancel out -a+b in both numerator and denominator.

$$-\left(-1\right)-1$$

Combine -\left(-1\right) and -1 to get 0.

$$0$$

Factor

$$0$$