What is the answer to solvefor y,x^{2/3}+y^{2/3}=1 ?

The answer to solvefor y,x^{2/3}+y^{2/3}=1 is

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solvefor y,x^{2/3}+y^{2/3}=1

solve for

y, x^{\frac{2}{3}} + y^{\frac{2}{3}} = 1

y = 1 - 3 x^{\frac{2}{3}} + 3 x^{\frac{4}{3}} - x^{2} {- 1 \leq x \leq 1}, y = - 1 - 3 x^{\frac{2}{3}} + 3 x^{\frac{4}{3}} - x^{2} {- 1 \leq x \leq 1}

Solution steps

x^{\frac{2}{3}} + y^{\frac{2}{3}} = 1

Move

x^{\frac{2}{3}}

to the right side

y^{\frac{2}{3}} = 1 - x^{\frac{2}{3}}

Take both sides of the equation to the power of

3 : y^{2} = 1 - 3 x^{\frac{2}{3}} + 3 x^{\frac{4}{3}} - x^{2}

y^{2} = 1 - 3 x^{\frac{2}{3}} + 3 x^{\frac{4}{3}} - x^{2}

Solve

y^{2} = 1 - 3 x^{\frac{2}{3}} + 3 x^{\frac{4}{3}} - x^{2} : y = 1 - 3 x^{\frac{2}{3}} + 3 x^{\frac{4}{3}} - x^{2}, y = - 1 - 3 x^{\frac{2}{3}} + 3 x^{\frac{4}{3}} - x^{2}

y = 1 - 3 x^{\frac{2}{3}} + 3 x^{\frac{4}{3}} - x^{2}, y = - 1 - 3 x^{\frac{2}{3}} + 3 x^{\frac{4}{3}} - x^{2}

Verify Solutions:

y = 1 - 3 x^{\frac{2}{3}} + 3 x^{\frac{4}{3}} - x^{2} {- 1 \leq x \leq 1}, y = - 1 - 3 x^{\frac{2}{3}} + 3 x^{\frac{4}{3}} - x^{2} {- 1 \leq x \leq 1}

The solutions are

y = 1 - 3 x^{\frac{2}{3}} + 3 x^{\frac{4}{3}} - x^{2} {- 1 \leq x \leq 1}, y = - 1 - 3 x^{\frac{2}{3}} + 3 x^{\frac{4}{3}} - x^{2} {- 1 \leq x \leq 1}

1 - \frac{1}{3 x ^{\frac{2}{3}}} = 0

y, x = 4 - y^{2}

2 x - 6 - x - 5 = x - 1

2 x + 8 = 3 x - 2

2 x + 8 = - 5

What is the answer to solvefor y,x^{2/3}+y^{2/3}=1 ?
The answer to solvefor y,x^{2/3}+y^{2/3}=1 is