What is the answer to solvefor y,x=\sqrt[3]{(-8y^3)/(y^3-1)} ?

The answer to solvefor y,x=\sqrt[3]{(-8y^3)/(y^3-1)} is y= x/(\sqrt[3]{x^3+8)} {x -2}

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solvefor y,x=\sqrt[3]{(-8y^3)/(y^3-1)}

solve for

y, x = 3 \frac{- 8 y ^{3}}{y ^{3} - 1}

y = \frac{x}{3 x ^{3} + 8} {x < - 2 or x > - 2}

Solution steps

x = 3 \frac{- 8 y ^{3}}{y ^{3} - 1}

Take both sides of the equation to the power of

3 : x^{3} = - \frac{8 y ^{3}}{y ^{3} - 1}

x^{3} = - \frac{8 y ^{3}}{y ^{3} - 1}

Solve

x^{3} = - \frac{8 y ^{3}}{y ^{3} - 1} : y = \frac{x}{3 x ^{3} + 8}

y = \frac{x}{3 x ^{3} + 8}

Verify Solutions:

y = \frac{x}{3 x ^{3} + 8} {x < - 2 or x > - 2}

The solution is

y = \frac{x}{3 x ^{3} + 8} {x < - 2 or x > - 2}

x + 78 = x + 6

8 3 10 x - 15 = 17

3 2 x - 4 = - 2

6 = x - 2

6 x - 5 + 10 = 3

What is the answer to solvefor y,x=\sqrt[3]{(-8y^3)/(y^3-1)} ?
The answer to solvefor y,x=\sqrt[3]{(-8y^3)/(y^3-1)} is y= x/(\sqrt[3]{x^3+8)} {x<-2\lor x>-2}