What is the answer to solvefor a,sqrt(a(b+1))=((1))/((b^2)) ?

The answer to solvefor a,sqrt(a(b+1))=((1))/((b^2)) is a= 1/(b^4(b+1)) {b 0}

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solvefor a,sqrt(a(b+1))=((1))/((b^2))

solve for

a, a (b + 1) = \frac{( 1 )}{( b ^{2} )}

a = \frac{1}{b ^{4} ( b + 1 )} {b < - 1 or - 1 < b < 0 or b > 0}

Solution steps

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

Square both sides:

ab + a = \frac{1}{b ^{4}}

ab + a = \frac{1}{b ^{4}}

Solve

ab + a = \frac{1}{b ^{4}} : a = \frac{1}{b ^{4} ( b + 1 )}

a = \frac{1}{b ^{4} ( b + 1 )}

Verify Solutions:

a = \frac{1}{b ^{4} ( b + 1 )} {b < - 1 or - 1 < b < 0 or b > 0}

The solution is

a = \frac{1}{b ^{4} ( b + 1 )} {b < - 1 or - 1 < b < 0 or b > 0}

(6 x + 9)^{\frac{1}{3}} + 8 = 3

(v - 8)^{\frac{3}{2}} = 2

\frac{2 x + 5 - 3}{( 2 x + 5 )} = - 2

c \cdot x^{2} + x + 1 = 1

x + x = 6

What is the answer to solvefor a,sqrt(a(b+1))=((1))/((b^2)) ?
The answer to solvefor a,sqrt(a(b+1))=((1))/((b^2)) is a= 1/(b^4(b+1)) {b<-1\lor-1<b<0\lor b>0}