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

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

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

solve for

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

y = 8 - 12 x^{\frac{2}{3}} + 6 x^{\frac{4}{3}} - x^{2} {- 2 \cdot 2^{\frac{1}{2}} < x < 2 \cdot 2^{\frac{1}{2}}}, y = - 8 - 12 x^{\frac{2}{3}} + 6 x^{\frac{4}{3}} - x^{2} {- 2 \cdot 2^{\frac{1}{2}} < x < 2 \cdot 2^{\frac{1}{2}}}

Solution steps

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

Move

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

to the right side

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

Take both sides of the equation to the power of

3 : y^{2} = 8 - 12 x^{\frac{2}{3}} + 6 x^{\frac{4}{3}} - x^{2}

y^{2} = 8 - 12 x^{\frac{2}{3}} + 6 x^{\frac{4}{3}} - x^{2}

Solve

y^{2} = 8 - 12 x^{\frac{2}{3}} + 6 x^{\frac{4}{3}} - x^{2} : y = 8 - 12 x^{\frac{2}{3}} + 6 x^{\frac{4}{3}} - x^{2}, y = - 8 - 12 x^{\frac{2}{3}} + 6 x^{\frac{4}{3}} - x^{2}

y = 8 - 12 x^{\frac{2}{3}} + 6 x^{\frac{4}{3}} - x^{2}, y = - 8 - 12 x^{\frac{2}{3}} + 6 x^{\frac{4}{3}} - x^{2}

Verify Solutions:

y = 8 - 12 x^{\frac{2}{3}} + 6 x^{\frac{4}{3}} - x^{2} {- 2 \cdot 2^{\frac{1}{2}} < x < 2 \cdot 2^{\frac{1}{2}}}, y = - 8 - 12 x^{\frac{2}{3}} + 6 x^{\frac{4}{3}} - x^{2} {- 2 \cdot 2^{\frac{1}{2}} < x < 2 \cdot 2^{\frac{1}{2}}}

The solutions are

y = 8 - 12 x^{\frac{2}{3}} + 6 x^{\frac{4}{3}} - x^{2} {- 2 \cdot 2^{\frac{1}{2}} < x < 2 \cdot 2^{\frac{1}{2}}}, y = - 8 - 12 x^{\frac{2}{3}} + 6 x^{\frac{4}{3}} - x^{2} {- 2 \cdot 2^{\frac{1}{2}} < x < 2 \cdot 2^{\frac{1}{2}}}

2 x + 3 = x - 1

(3 x - 5)^{\frac{1}{4}} = 2

y, x = 1 - y

\frac{3 x + 18}{2 x + 9} = 0

x = x + 110

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