What is ((1/a+1/((b+c))))/((1/a-1/((b+c)))) ?

The solution to ((1/a+1/((b+c))))/((1/a-1/((b+c)))) is (b+c+a)/(b+c-a)

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simplify ((1/a+1/((b+c))))/((1/a-1/((b+c))))

simplify

\frac{( \frac{1}{a} + \frac{1}{( b + c )} )}{( \frac{1}{a} - \frac{1}{( b + c )} )}

\frac{b + c + a}{b + c - a}

Solution steps

\frac{\frac{1}{a} + \frac{1}{b + c}}{\frac{1}{a} - \frac{1}{b + c}}

Join

\frac{1}{a} + \frac{1}{b + c} : \frac{b + c + a}{a ( b + c )}

= \frac{\frac{b + c + a}{a ( b + c )}}{\frac{1}{a} - \frac{1}{b + c}}

Join

\frac{1}{a} - \frac{1}{b + c} : \frac{b + c - a}{a ( b + c )}

= \frac{\frac{b + c + a}{a ( b + c )}}{\frac{b + c - a}{a ( b + c )}}

Apply the fraction rule:

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

= \frac{( b + c + a ) a ( b + c )}{a ( b + c ) ( b + c - a )}

Cancel

\frac{( b + c + a ) a ( b + c )}{a ( b + c ) ( b + c - a )} : \frac{b + c + a}{b + c - a}

= \frac{b + c + a}{b + c - a}

4 x^{3} - 2 x^{2} + x - 3

(4 m^{4} + 81 n^{u})^{3}

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

p^{3} - 343

(3 b + 6 a)^{3}

What is ((1/a+1/((b+c))))/((1/a-1/((b+c)))) ?
The solution to ((1/a+1/((b+c))))/((1/a-1/((b+c)))) is (b+c+a)/(b+c-a)