Solve 4p/(2p-2q)^2-q/(q-p)^2

$$\frac { 4 p } { ( 2 p - 2 q ) ^ { 2 } } - \frac { q } { ( q - p ) ^ { 2 } }$$

Evaluate

$$-\frac{1}{q-p}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(2p-2q\right)^{2} and \left(q-p\right)^{2} is \left(-p+q\right)^{2}\left(2p-2q\right)^{2}. Multiply \frac{4p}{\left(2p-2q\right)^{2}} times \frac{\left(-p+q\right)^{2}}{\left(-p+q\right)^{2}}. Multiply \frac{q}{\left(q-p\right)^{2}} times \frac{\left(2p-2q\right)^{2}}{\left(2p-2q\right)^{2}}.

$$\frac{4p\left(-p+q\right)^{2}}{\left(-p+q\right)^{2}\left(2p-2q\right)^{2}}-\frac{q\left(2p-2q\right)^{2}}{\left(-p+q\right)^{2}\left(2p-2q\right)^{2}}$$

Since \frac{4p\left(-p+q\right)^{2}}{\left(-p+q\right)^{2}\left(2p-2q\right)^{2}} and \frac{q\left(2p-2q\right)^{2}}{\left(-p+q\right)^{2}\left(2p-2q\right)^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{4p\left(-p+q\right)^{2}-q\left(2p-2q\right)^{2}}{\left(-p+q\right)^{2}\left(2p-2q\right)^{2}}$$

Do the multiplications in 4p\left(-p+q\right)^{2}-q\left(2p-2q\right)^{2}.

$$\frac{4p^{3}-8p^{2}q+4pq^{2}-4qp^{2}+8q^{2}p-4q^{3}}{\left(-p+q\right)^{2}\left(2p-2q\right)^{2}}$$

Combine like terms in 4p^{3}-8p^{2}q+4pq^{2}-4qp^{2}+8q^{2}p-4q^{3}.

$$\frac{4p^{3}+12pq^{2}-12p^{2}q-4q^{3}}{\left(-p+q\right)^{2}\left(2p-2q\right)^{2}}$$

Expand \left(-p+q\right)^{2}\left(2p-2q\right)^{2}.

$$\frac{4p^{3}+12pq^{2}-12p^{2}q-4q^{3}}{4p^{4}-16pq^{3}+4q^{4}-16qp^{3}+24\left(pq\right)^{2}}$$

Expand \left(pq\right)^{2}.

$$\frac{4p^{3}+12pq^{2}-12p^{2}q-4q^{3}}{4p^{4}-16pq^{3}+4q^{4}-16qp^{3}+24p^{2}q^{2}}$$

Factor the expressions that are not already factored.

$$\frac{4\left(p-q\right)^{3}}{4\left(p-q\right)^{4}}$$

Cancel out 4\left(p-q\right)^{3} in both numerator and denominator.

$$\frac{1}{p-q}$$

Expand

$$-\frac{1}{q-p}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(2p-2q\right)^{2} and \left(q-p\right)^{2} is \left(-p+q\right)^{2}\left(2p-2q\right)^{2}. Multiply \frac{4p}{\left(2p-2q\right)^{2}} times \frac{\left(-p+q\right)^{2}}{\left(-p+q\right)^{2}}. Multiply \frac{q}{\left(q-p\right)^{2}} times \frac{\left(2p-2q\right)^{2}}{\left(2p-2q\right)^{2}}.

$$\frac{4p\left(-p+q\right)^{2}}{\left(-p+q\right)^{2}\left(2p-2q\right)^{2}}-\frac{q\left(2p-2q\right)^{2}}{\left(-p+q\right)^{2}\left(2p-2q\right)^{2}}$$

Since \frac{4p\left(-p+q\right)^{2}}{\left(-p+q\right)^{2}\left(2p-2q\right)^{2}} and \frac{q\left(2p-2q\right)^{2}}{\left(-p+q\right)^{2}\left(2p-2q\right)^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{4p\left(-p+q\right)^{2}-q\left(2p-2q\right)^{2}}{\left(-p+q\right)^{2}\left(2p-2q\right)^{2}}$$

Do the multiplications in 4p\left(-p+q\right)^{2}-q\left(2p-2q\right)^{2}.

$$\frac{4p^{3}-8p^{2}q+4pq^{2}-4qp^{2}+8q^{2}p-4q^{3}}{\left(-p+q\right)^{2}\left(2p-2q\right)^{2}}$$

Combine like terms in 4p^{3}-8p^{2}q+4pq^{2}-4qp^{2}+8q^{2}p-4q^{3}.

$$\frac{4p^{3}+12pq^{2}-12p^{2}q-4q^{3}}{\left(-p+q\right)^{2}\left(2p-2q\right)^{2}}$$

Expand \left(-p+q\right)^{2}\left(2p-2q\right)^{2}.

$$\frac{4p^{3}+12pq^{2}-12p^{2}q-4q^{3}}{4p^{4}-16pq^{3}+4q^{4}-16qp^{3}+24\left(pq\right)^{2}}$$

Expand \left(pq\right)^{2}.

$$\frac{4p^{3}+12pq^{2}-12p^{2}q-4q^{3}}{4p^{4}-16pq^{3}+4q^{4}-16qp^{3}+24p^{2}q^{2}}$$

Factor the expressions that are not already factored.

$$\frac{4\left(p-q\right)^{3}}{4\left(p-q\right)^{4}}$$

Cancel out 4\left(p-q\right)^{3} in both numerator and denominator.

$$\frac{1}{p-q}$$