What is 1/(p^2-q^2)*(p+q)/p ?

The solution to 1/(p^2-q^2)*(p+q)/p is 1/((p-q)p)

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simplify 1/(p^2-q^2)*(p+q)/p

simplify

\frac{1}{p ^{2} - q ^{2}} \cdot \frac{p + q}{p}

\frac{1}{( p - q ) p}

Solution steps

\frac{1}{p ^{2} - q ^{2}} \cdot \frac{p + q}{p}

Apply the fraction rule:

\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}

= \frac{1 \cdot ( p + q )}{( p ^{2} - q ^{2} ) p}

Apply rule:

1 \cdot a = a

1 \cdot (p + q) = (p + q)

= \frac{p + q}{( p ^{2} - q ^{2} ) p}

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

p^{2} - q^{2} = (p + q) (p - q)

= \frac{p + q}{( p + q ) ( p - q ) p}

Cancel the common factor:

p + q

= \frac{1}{( p - q ) p}

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