Solve a^2(a-1)x=a^2+a+2ax

$$a ^ { 2 } ( a - 1 ) x = a ^ { 2 } + a + 2 a x$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=\frac{1}{a-2}\text{, }&a\neq 2\\x\in \mathrm{C}\text{, }&a=-1\text{ or }a=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply a^{2} by a-1.

$$\left(a^{3}-a^{2}\right)x=a^{2}+a+2ax$$

Use the distributive property to multiply a^{3}-a^{2} by x.

$$a^{3}x-a^{2}x=a^{2}+a+2ax$$

Subtract 2ax from both sides.

$$a^{3}x-a^{2}x-2ax=a^{2}+a$$

Combine all terms containing x.

$$\left(a^{3}-a^{2}-2a\right)x=a^{2}+a$$

Divide both sides by a^{3}-a^{2}-2a.

$$\frac{\left(a^{3}-a^{2}-2a\right)x}{a^{3}-a^{2}-2a}=\frac{a\left(a+1\right)}{a^{3}-a^{2}-2a}$$

Dividing by a^{3}-a^{2}-2a undoes the multiplication by a^{3}-a^{2}-2a.

$$x=\frac{a\left(a+1\right)}{a^{3}-a^{2}-2a}$$

Divide a\left(1+a\right) by a^{3}-a^{2}-2a.

$$x=\frac{1}{a-2}$$

Solve for x

$$\left\{\begin{matrix}x=\frac{1}{a-2}\text{, }&a\neq 2\\x\in \mathrm{R}\text{, }&a=-1\text{ or }a=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply a^{2} by a-1.

$$\left(a^{3}-a^{2}\right)x=a^{2}+a+2ax$$

Use the distributive property to multiply a^{3}-a^{2} by x.

$$a^{3}x-a^{2}x=a^{2}+a+2ax$$

Subtract 2ax from both sides.

$$a^{3}x-a^{2}x-2ax=a^{2}+a$$

Combine all terms containing x.

$$\left(a^{3}-a^{2}-2a\right)x=a^{2}+a$$

Divide both sides by a^{3}-a^{2}-2a.

$$\frac{\left(a^{3}-a^{2}-2a\right)x}{a^{3}-a^{2}-2a}=\frac{a\left(a+1\right)}{a^{3}-a^{2}-2a}$$

Dividing by a^{3}-a^{2}-2a undoes the multiplication by a^{3}-a^{2}-2a.

$$x=\frac{a\left(a+1\right)}{a^{3}-a^{2}-2a}$$

Divide a\left(1+a\right) by a^{3}-a^{2}-2a.

$$x=\frac{1}{a-2}$$

Solve for a

$$\left\{\begin{matrix}\\a=-1\text{; }a=0\text{, }&\text{unconditionally}\\a=2+\frac{1}{x}\text{, }&x\neq 0\end{matrix}\right.$$