What is the answer to solvefor x,x^{2/3}+y^{2/3}=4 ?

The answer to solvefor x,x^{2/3}+y^{2/3}=4 is

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solvefor x,x^{2/3}+y^{2/3}=4

solve for

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

x = 64 - 48 y^{\frac{2}{3}} + 12 y^{\frac{4}{3}} - y^{2} {- 8 \leq y \leq 8}, x = - 64 - 48 y^{\frac{2}{3}} + 12 y^{\frac{4}{3}} - y^{2} {- 8 \leq y \leq 8}

Solution steps

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

Move

y^{\frac{2}{3}}

to the right side

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

Take both sides of the equation to the power of

3 : x^{2} = 64 - 48 y^{\frac{2}{3}} + 12 y^{\frac{4}{3}} - y^{2}

x^{2} = 64 - 48 y^{\frac{2}{3}} + 12 y^{\frac{4}{3}} - y^{2}

Solve

x^{2} = 64 - 48 y^{\frac{2}{3}} + 12 y^{\frac{4}{3}} - y^{2} : x = 64 - 48 y^{\frac{2}{3}} + 12 y^{\frac{4}{3}} - y^{2}, x = - 64 - 48 y^{\frac{2}{3}} + 12 y^{\frac{4}{3}} - y^{2}

x = 64 - 48 y^{\frac{2}{3}} + 12 y^{\frac{4}{3}} - y^{2}, x = - 64 - 48 y^{\frac{2}{3}} + 12 y^{\frac{4}{3}} - y^{2}

Verify Solutions:

x = 64 - 48 y^{\frac{2}{3}} + 12 y^{\frac{4}{3}} - y^{2} {- 8 \leq y \leq 8}, x = - 64 - 48 y^{\frac{2}{3}} + 12 y^{\frac{4}{3}} - y^{2} {- 8 \leq y \leq 8}

The solutions are

x = 64 - 48 y^{\frac{2}{3}} + 12 y^{\frac{4}{3}} - y^{2} {- 8 \leq y \leq 8}, x = - 64 - 48 y^{\frac{2}{3}} + 12 y^{\frac{4}{3}} - y^{2} {- 8 \leq y \leq 8}

4 \cdot (x + 5)^{2} = 6

x, x^{\frac{2}{3}} + y^{\frac{2}{3}} = 7

\frac{a}{5 - 2 x + 5} = 3

x, z = x \cdot (x^{2} + y^{2})^{- \frac{1}{2}}

2 x + 5 = x + 2

What is the answer to solvefor x,x^{2/3}+y^{2/3}=4 ?
The answer to solvefor x,x^{2/3}+y^{2/3}=4 is