Solve -x^3+Ax^2=B

$$- x ^ { 3 } + A x ^ { 2 } = B$$

Solve for A (complex solution)

$$\left\{\begin{matrix}A=x+\frac{B}{x^{2}}\text{, }&x\neq 0\\A\in \mathrm{C}\text{, }&B=0\text{ and }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Add x^{3} to both sides.

$$Ax^{2}=B+x^{3}$$

The equation is in standard form.

$$x^{2}A=x^{3}+B$$

Divide both sides by x^{2}.

$$\frac{x^{2}A}{x^{2}}=\frac{x^{3}+B}{x^{2}}$$

Dividing by x^{2} undoes the multiplication by x^{2}.

$$A=\frac{x^{3}+B}{x^{2}}$$

Divide B+x^{3} by x^{2}.

$$A=x+\frac{B}{x^{2}}$$

Solve for A

$$\left\{\begin{matrix}A=x+\frac{B}{x^{2}}\text{, }&x\neq 0\\A\in \mathrm{R}\text{, }&x=0\text{ and }B=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Add x^{3} to both sides.

$$Ax^{2}=B+x^{3}$$

The equation is in standard form.

$$x^{2}A=x^{3}+B$$

Divide both sides by x^{2}.

$$\frac{x^{2}A}{x^{2}}=\frac{x^{3}+B}{x^{2}}$$

Dividing by x^{2} undoes the multiplication by x^{2}.

$$A=\frac{x^{3}+B}{x^{2}}$$

Divide B+x^{3} by x^{2}.

$$A=x+\frac{B}{x^{2}}$$

Solve for B

$$B=\left(A-x\right)x^{2}$$