What is the answer to y^9\times (y^2)^{-1}y=-2 ?

The answer to y^9\times (y^2)^{-1}y=-2 is

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y^9\times (y^2)^{-1}y=-2

y^{9} \cdot (y^{2})^{- 1} y = - 2

y = \frac{1}{2 ^{\frac{5}{8}} 2 - 1} + \frac{2 - 1}{2 ^{\frac{5}{8}}} i, y = - \frac{1}{2 ^{\frac{5}{4}} \cdot \frac{2 - 1}{2 ^{\frac{5}{8}}}} - \frac{2 - 1}{2 ^{\frac{5}{8}}} i, y = - \frac{1}{2 ^{\frac{5}{8}} 1 + 2} + \frac{1 + 2}{2 ^{\frac{5}{8}}} i, y = \frac{1}{2 ^{\frac{5}{8}} 1 + 2} - \frac{1 + 2}{2 ^{\frac{5}{8}}} i, y = \frac{1}{2 ^{\frac{5}{8}} 1 + 2} + \frac{1 + 2}{2 ^{\frac{5}{8}}} i, y = - \frac{1}{2 ^{\frac{5}{4}} \cdot \frac{1 + 2}{2 ^{\frac{5}{8}}}} - \frac{1 + 2}{2 ^{\frac{5}{8}}} i, y = - \frac{1}{2 ^{\frac{5}{8}} 2 - 1} + \frac{2 - 1}{2 ^{\frac{5}{8}}} i, y = \frac{1}{2 ^{\frac{5}{8}} 2 - 1} - \frac{2 - 1}{2 ^{\frac{5}{8}}} i

Solution steps

y^{9} (y^{2})^{- 1} y = - 2

Refine

y^{8} = - 2

Solve

y^{8} = - 2 : y = \frac{1}{2 ^{\frac{5}{8}} 2 - 1} + \frac{2 - 1}{2 ^{\frac{5}{8}}} i, y = - \frac{1}{2 ^{\frac{5}{4}} \cdot \frac{2 - 1}{2 ^{\frac{5}{8}}}} - \frac{2 - 1}{2 ^{\frac{5}{8}}} i, y = - \frac{1}{2 ^{\frac{5}{8}} 1 + 2} + \frac{1 + 2}{2 ^{\frac{5}{8}}} i, y = \frac{1}{2 ^{\frac{5}{8}} 1 + 2} - \frac{1 + 2}{2 ^{\frac{5}{8}}} i, y = \frac{1}{2 ^{\frac{5}{8}} 1 + 2} + \frac{1 + 2}{2 ^{\frac{5}{8}}} i, y = - \frac{1}{2 ^{\frac{5}{4}} \cdot \frac{1 + 2}{2 ^{\frac{5}{8}}}} - \frac{1 + 2}{2 ^{\frac{5}{8}}} i, y = - \frac{1}{2 ^{\frac{5}{8}} 2 - 1} + \frac{2 - 1}{2 ^{\frac{5}{8}}} i, y = \frac{1}{2 ^{\frac{5}{8}} 2 - 1} - \frac{2 - 1}{2 ^{\frac{5}{8}}} i

y = \frac{1}{2 ^{\frac{5}{8}} 2 - 1} + \frac{2 - 1}{2 ^{\frac{5}{8}}} i, y = - \frac{1}{2 ^{\frac{5}{4}} \cdot \frac{2 - 1}{2 ^{\frac{5}{8}}}} - \frac{2 - 1}{2 ^{\frac{5}{8}}} i, y = - \frac{1}{2 ^{\frac{5}{8}} 1 + 2} + \frac{1 + 2}{2 ^{\frac{5}{8}}} i, y = \frac{1}{2 ^{\frac{5}{8}} 1 + 2} - \frac{1 + 2}{2 ^{\frac{5}{8}}} i, y = \frac{1}{2 ^{\frac{5}{8}} 1 + 2} + \frac{1 + 2}{2 ^{\frac{5}{8}}} i, y = - \frac{1}{2 ^{\frac{5}{4}} \cdot \frac{1 + 2}{2 ^{\frac{5}{8}}}} - \frac{1 + 2}{2 ^{\frac{5}{8}}} i, y = - \frac{1}{2 ^{\frac{5}{8}} 2 - 1} + \frac{2 - 1}{2 ^{\frac{5}{8}}} i, y = \frac{1}{2 ^{\frac{5}{8}} 2 - 1} - \frac{2 - 1}{2 ^{\frac{5}{8}}} i

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What is the answer to y^9\times (y^2)^{-1}y=-2 ?
The answer to y^9\times (y^2)^{-1}y=-2 is
Solve for y: y^9\times (y^2)^{-1}y=-2
The solution is