What is the answer to solvefor y,x^3+3y-xy=0 ?

The answer to solvefor y,x^3+3y-xy=0 is y=-(x^3)/(3-x);\quad x\ne 3

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solvefor y,x^3+3y-xy=0

solve for

y, x^{3} + 3 y - x y = 0

y = - \frac{x ^{3}}{3 - x}; x \neq = 3

Solution steps

x^{3} + 3 y - x y = 0

Move

x^{3}

to the right side

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

Factor

3 y - x y : y (3 - x)

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

Divide both sides by

3 - x; x \neq = 3

y = - \frac{x ^{3}}{3 - x}; x \neq = 3

2 (x^{2} - x) - 2 x (x + 3) = 5 x - 1

x, 15 x + 20 y = 220

k = c (p - 5)

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

x, \frac{x}{2} + \frac{y}{3} = 8

What is the answer to solvefor y,x^3+3y-xy=0 ?
The answer to solvefor y,x^3+3y-xy=0 is y=-(x^3)/(3-x);\quad x\ne 3