Solve ac^2=ab^2+bc^2

$$a c ^ { 2 } = a b ^ { 2 } + b c ^ { 2 }$$

Solve for a

$$\left\{\begin{matrix}a=\frac{bc^{2}}{c^{2}-b^{2}}\text{, }&|c|\neq |b|\\a\in \mathrm{R}\text{, }&c=0\text{ and }b=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Subtract ab^{2} from both sides.

$$ac^{2}-ab^{2}=bc^{2}$$

Reorder the terms.

$$-ab^{2}+ac^{2}=bc^{2}$$

Combine all terms containing a.

$$\left(-b^{2}+c^{2}\right)a=bc^{2}$$

The equation is in standard form.

$$\left(c^{2}-b^{2}\right)a=bc^{2}$$

Divide both sides by -b^{2}+c^{2}.

$$\frac{\left(c^{2}-b^{2}\right)a}{c^{2}-b^{2}}=\frac{bc^{2}}{c^{2}-b^{2}}$$

Dividing by -b^{2}+c^{2} undoes the multiplication by -b^{2}+c^{2}.

$$a=\frac{bc^{2}}{c^{2}-b^{2}}$$

Solve for b

$$\left\{\begin{matrix}b=-\frac{c\left(\sqrt{4a^{2}+c^{2}}+c\right)}{2a}\text{; }b=-\frac{c\left(-\sqrt{4a^{2}+c^{2}}+c\right)}{2a}\text{, }&a\neq 0\\b=0\text{, }&a=0\text{ and }c\neq 0\\b\in \mathrm{R}\text{, }&a=0\text{ and }c=0\end{matrix}\right.$$