What are the solutions to the equation (m-4)^3=(1/8)^{-1} ?

The solutions to the equation (m-4)^3=(1/8)^{-1} are m=6,m=3+sqrt(3)i,m=3-sqrt(3)i

Find the zeros of (m-4)^3=(1/8)^{-1}

The zeros of (m-4)^3=(1/8)^{-1} are m=6,m=3+sqrt(3)i,m=3-sqrt(3)i

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(m - 4)^{3} = (\frac{1}{8})^{- 1}

m = 6, m = 3 + 3 i, m = 3 - 3 i

Solution steps

(m - 4)^{3} = (\frac{1}{8})^{- 1}

Simplify

(\frac{1}{8})^{- 1} : 8

For

g^{3} (x) = f (a)

the solutions are

Solve

The solutions are

m = 6, m = 3 + 3 i, m = 3 - 3 i

x^{4} - 3 x - 4 = 0

0 = 8 x^{3} - 125

5 y - 62 = y - 2 - y^{3}

12 x^{3} + 18 x^{2} - 18 x - 12 = 0

(5 y^{4})^{3} = 125 y^{24}

What are the solutions to the equation (m-4)^3=(1/8)^{-1} ?
The solutions to the equation (m-4)^3=(1/8)^{-1} are m=6,m=3+sqrt(3)i,m=3-sqrt(3)i
Find the zeros of (m-4)^3=(1/8)^{-1}
The zeros of (m-4)^3=(1/8)^{-1} are m=6,m=3+sqrt(3)i,m=3-sqrt(3)i