What are the solutions to the equation solvefor y,x^3y^2=2x^2+y^2 ?

The solutions to the equation solvefor y,x^3y^2=2x^2+y^2 are

Find the zeros of solvefor y,x^3y^2=2x^2+y^2

The zeros of solvefor y,x^3y^2=2x^2+y^2 are

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solve for

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

y = \frac{2 x}{x ^{3} - 1}, y = - \frac{2 x}{x ^{3} - 1}; x \neq = 1, x \neq = - \frac{1}{2} + \frac{3}{2} i, x \neq = - \frac{1}{2} - \frac{3}{2} i

Solution steps

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

Move

y^{2}

to the left side

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

Factor

x^{3} y^{2} - y^{2} : y^{2} (x^{3} - 1)

y^{2} (x^{3} - 1) = 2 x^{2}

Divide both sides by

x^{3} - 1; x \neq = 1, x \neq = - \frac{1}{2} + \frac{3}{2} i, x \neq = - \frac{1}{2} - \frac{3}{2} i

y^{2} = \frac{2 x ^{2}}{x ^{3} - 1}; x \neq = 1, x \neq = - \frac{1}{2} + \frac{3}{2} i, x \neq = - \frac{1}{2} - \frac{3}{2} i

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

y = \frac{2 x ^{2}}{x ^{3} - 1}, y = - \frac{2 x ^{2}}{x ^{3} - 1}; x \neq = 1, x \neq = - \frac{1}{2} + \frac{3}{2} i, x \neq = - \frac{1}{2} - \frac{3}{2} i

Simplify

\frac{2 x ^{2}}{x ^{3} - 1} : \frac{2 x}{x ^{3} - 1}

Simplify

- \frac{2 x ^{2}}{x ^{3} - 1} : - \frac{2 x}{x ^{3} - 1}

y = \frac{2 x}{x ^{3} - 1}, y = - \frac{2 x}{x ^{3} - 1}; x \neq = 1, x \neq = - \frac{1}{2} + \frac{3}{2} i, x \neq = - \frac{1}{2} - \frac{3}{2} i

3 x^{2} - 12 x + 1

4 y^{2} + 9 - 12 y = 5

x, x^{2} - 2 x + 2 = 0

y, f = x^{23} - 10 y + 15 x y^{2}

x^{2} - 19 = 0

What are the solutions to the equation solvefor y,x^3y^2=2x^2+y^2 ?
The solutions to the equation solvefor y,x^3y^2=2x^2+y^2 are
Find the zeros of solvefor y,x^3y^2=2x^2+y^2
The zeros of solvefor y,x^3y^2=2x^2+y^2 are