What are the solutions to the equation solvefor x,(x^2+y^2-1)=x^2y^3 ?

The solutions to the equation solvefor x,(x^2+y^2-1)=x^2y^3 are

Find the zeros of solvefor x,(x^2+y^2-1)=x^2y^3

The zeros of solvefor x,(x^2+y^2-1)=x^2y^3 are

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solve for

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

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

Solution steps

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

Refine

x^{2} + y^{2} - 1 = x^{2} y^{3}

Move

y^{2}

to the right side

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

Move

1

to the right side

x^{2} = x^{2} y^{3} - y^{2} + 1

Move

x^{2} y^{3}

to the left side

x^{2} - x^{2} y^{3} = - y^{2} + 1

Factor

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

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

Divide both sides by

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

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

For

x^{2} = f (a)

the solutions are

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

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

x, t (x) = 18 - 24 x + 2 x^{2}

9 x^{2} - 50 x + 57 = 0

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

- 10 z^{2} - 5 z = 0

7 x^{2} - 5 x + 6 = - 8

What are the solutions to the equation solvefor x,(x^2+y^2-1)=x^2y^3 ?
The solutions to the equation solvefor x,(x^2+y^2-1)=x^2y^3 are
Find the zeros of solvefor x,(x^2+y^2-1)=x^2y^3
The zeros of solvefor x,(x^2+y^2-1)=x^2y^3 are