What is the answer to x^{-6}-3x^{-3}-28=0 ?

The answer to x^{-6}-3x^{-3}-28=0 is

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x^{-6}-3x^{-3}-28=0

x^{- 6} - 3 x^{- 3} - 28 = 0

x = - \frac{1}{3 4}, x = \frac{1}{2 3 4} - \frac{3}{2 3 4} i, x = \frac{1}{2 3 4} + \frac{3}{2 3 4} i, x = \frac{1}{3 7}, x = - \frac{1}{2 3 7} + \frac{3}{2 3 7} i, x = - \frac{1}{2 3 7} - \frac{3}{2 3 7} i

Solution steps

x^{- 6} - 3 x^{- 3} - 28 = 0

Refine

\frac{1}{x ^{6}} - \frac{3}{x ^{3}} - 28 = 0

Multiply by LCM

1 - 3 x^{3} - 28 x^{6} = 0

Solve

1 - 3 x^{3} - 28 x^{6} = 0 : x = - \frac{1}{3 4}, x = \frac{1}{2 3 4} - \frac{3}{2 3 4} i, x = \frac{1}{2 3 4} + \frac{3}{2 3 4} i, x = \frac{1}{3 7}, x = - \frac{1}{2 3 7} + \frac{3}{2 3 7} i, x = - \frac{1}{2 3 7} - \frac{3}{2 3 7} i

x = - \frac{1}{3 4}, x = \frac{1}{2 3 4} - \frac{3}{2 3 4} i, x = \frac{1}{2 3 4} + \frac{3}{2 3 4} i, x = \frac{1}{3 7}, x = - \frac{1}{2 3 7} + \frac{3}{2 3 7} i, x = - \frac{1}{2 3 7} - \frac{3}{2 3 7} i

Verify Solutions

Find undefined (singularity) points:

x = 0

Combine undefined points with solutions:

x = - \frac{1}{3 4}, x = \frac{1}{2 3 4} - \frac{3}{2 3 4} i, x = \frac{1}{2 3 4} + \frac{3}{2 3 4} i, x = \frac{1}{3 7}, x = - \frac{1}{2 3 7} + \frac{3}{2 3 7} i, x = - \frac{1}{2 3 7} - \frac{3}{2 3 7} i

\frac{150 \cdot t}{t - 1} = 160

- \frac{8}{x - 4} + 1 = - \frac{x}{x - 4}

\frac{1}{10} = \frac{1}{20} + \frac{1}{x}

\frac{1}{6} + \frac{1}{3 x} = \frac{3}{2 x}

- \frac{1}{x ^{2}} = - \frac{1}{5}

What is the answer to x^{-6}-3x^{-3}-28=0 ?
The answer to x^{-6}-3x^{-3}-28=0 is
Solve for x: x^{-6}-3x^{-3}-28=0
The solution is