What is the answer to solvefor y,x=\sqrt[3]{y^2} ?

The answer to solvefor y,x=\sqrt[3]{y^2} is y=xsqrt(x) {x>= 0},y=-xsqrt(x) {x>= 0}

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solvefor y,x=\sqrt[3]{y^2}

solve for

y = x x {x \geq 0}, y = - x x {x \geq 0}

Solution steps

Take both sides of the equation to the power of

3 : x^{3} = y^{2}

x^{3} = y^{2}

Solve

x^{3} = y^{2} : y = x x, y = - x x

y = x x, y = - x x

Verify Solutions:

y = x x {x \geq 0}, y = - x x {x \geq 0}

The solutions are

y = x x {x \geq 0}, y = - x x {x \geq 0}

(x^{2} - 33)^{\frac{1}{2}} = 4

(2 x - 6)^{2} = 6 - 2 x

x + x = 380

x^{2} + 4 x - 1 = 2 x

What is the answer to solvefor y,x=\sqrt[3]{y^2} ?
The answer to solvefor y,x=\sqrt[3]{y^2} is y=xsqrt(x) {x>= 0},y=-xsqrt(x) {x>= 0}