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

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

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solvefor y,x=\sqrt[3]{y-1}

solve for

y = x^{3} + 1

Solution steps

Take both sides of the equation to the power of

3 : x^{3} = y - 1

x^{3} = y - 1

Solve

x^{3} = y - 1 : y = x^{3} + 1

y = x^{3} + 1

Verify Solutions:

y = x^{3} + 1

True

The solution is

y = x^{3} + 1

(x - 5)^{\frac{1}{2}} + 5 = 2

512 = 4 (x + 15)^{\frac{7}{6}}

2 x - \frac{4}{2 x - 4} - 4 = 0

x - 3 x = - 2

x - 2 + x - 13 = 11

What is the answer to solvefor y,x=\sqrt[3]{y-1} ?
The answer to solvefor y,x=\sqrt[3]{y-1} is y=x^3+1