What is the answer to b^{-3}=27 ?

The answer to b^{-3}=27 is b= 1/3 ,b=-1/6+1/(2sqrt(3))i,b=-1/6-1/(2sqrt(3))i

Solve for b: b^{-3}=27

The solution is b= 1/3 ,b=-1/6+1/(2sqrt(3))i,b=-1/6-1/(2sqrt(3))i

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b^{- 3} = 27

b = \frac{1}{3}, b = - \frac{1}{6} + \frac{1}{2 3} i, b = - \frac{1}{6} - \frac{1}{2 3} i

Solution steps

b^{- 3} = 27

Refine

\frac{1}{b ^{3}} = 27

Multiply both sides by

b^{3}

1 = 27 b^{3}

Solve

1 = 27 b^{3} : b = \frac{1}{3}, b = - \frac{1}{6} + \frac{1}{2 3} i, b = - \frac{1}{6} - \frac{1}{2 3} i

b = \frac{1}{3}, b = - \frac{1}{6} + \frac{1}{2 3} i, b = - \frac{1}{6} - \frac{1}{2 3} i

Verify Solutions

Find undefined (singularity) points:

b = 0

Combine undefined points with solutions:

b = \frac{1}{3}, b = - \frac{1}{6} + \frac{1}{2 3} i, b = - \frac{1}{6} - \frac{1}{2 3} i

\frac{100}{1 + x ^{2} + 2 y ^{2}} = 0

\frac{5 π}{4} = \frac{2 π}{b}

10 = 5 + \frac{80000}{x}

- \frac{9}{8 x + 19} = 0

\frac{3 c + 2}{( 3 c + 4 ) ( 3 c + 2 )} = 0

What is the answer to b^{-3}=27 ?
The answer to b^{-3}=27 is b= 1/3 ,b=-1/6+1/(2sqrt(3))i,b=-1/6-1/(2sqrt(3))i
Solve for b: b^{-3}=27
The solution is b= 1/3 ,b=-1/6+1/(2sqrt(3))i,b=-1/6-1/(2sqrt(3))i