What is (\frac{c^2-4)/(c+3)}{(c+2)/(3(c^2-9))} ?

The solution to (\frac{c^2-4)/(c+3)}{(c+2)/(3(c^2-9))} is 3(c-2)(c-3)

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simplify (\frac{c^2-4)/(c+3)}{(c+2)/(3(c^2-9))}

simplify

\frac{\frac{c ^{2} - 4}{c + 3}}{\frac{c + 2}{3 ( c ^{2} - 9 )}}

3 (c - 2) (c - 3)

Solution steps

\frac{\frac{c ^{2} - 4}{c + 3}}{\frac{c + 2}{3 ( c ^{2} - 9 )}}

Apply the fraction rule:

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

= \frac{( c ^{2} - 4 ) \cdot 3 ( c ^{2} - 9 )}{( c + 3 ) ( c + 2 )}

Cancel

\frac{( c ^{2} - 4 ) \cdot 3 ( c ^{2} - 9 )}{( c + 3 ) ( c + 2 )} : (c - 2) \cdot 3 (c - 3)

= (c - 2) \cdot 3 (c - 3)

= 3 (c - 2) (c - 3)

64 m^{9} + 64 n^{16}

\frac{c ^{2} + 7 c + 10}{c ^{2} + 2 c - 15} \cdot \frac{4 c + 12}{3 c + 15}

12 x^{2} - 4 x

(\frac{1 + cos ( 2 x )}{2})^{2}

(n + 5 k)^{3}

What is (\frac{c^2-4)/(c+3)}{(c+2)/(3(c^2-9))} ?
The solution to (\frac{c^2-4)/(c+3)}{(c+2)/(3(c^2-9))} is 3(c-2)(c-3)