Solve x^4-1/3x^3-y=0

$$x ^ { 4 } - \frac { 1 } { 3 } x ^ { 3 } - y = 0$$

Solve for y

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

Steps for Solving Linear Equation

Subtract x^{4} from both sides. Anything subtracted from zero gives its negation.

$$-\frac{1}{3}x^{3}-y=-x^{4}$$

Add \frac{1}{3}x^{3} to both sides.

$$-y=-x^{4}+\frac{1}{3}x^{3}$$

The equation is in standard form.

$$-y=-x^{4}+\frac{x^{3}}{3}$$

Divide both sides by -1.

$$\frac{-y}{-1}=\frac{\left(1-3x\right)x^{3}}{-3}$$

Dividing by -1 undoes the multiplication by -1.

$$y=\frac{\left(1-3x\right)x^{3}}{-3}$$

Divide \frac{\left(1-3x\right)x^{3}}{3} by -1.

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