Solve (x)=cx^2(x+2)^3(x-1)

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

Solve for c

$$\left\{\begin{matrix}c=\frac{1}{x\left(x-1\right)\left(x+2\right)^{3}}\text{, }&x\neq -2\text{ and }x\neq 1\text{ and }x\neq 0\\c\in \mathrm{R}\text{, }&x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} to expand \left(x+2\right)^{3}.

$$x=cx^{2}\left(x^{3}+6x^{2}+12x+8\right)\left(x-1\right)$$

Use the distributive property to multiply cx^{2} by x^{3}+6x^{2}+12x+8.

$$x=\left(cx^{5}+6cx^{4}+12cx^{3}+8cx^{2}\right)\left(x-1\right)$$

Use the distributive property to multiply cx^{5}+6cx^{4}+12cx^{3}+8cx^{2} by x-1 and combine like terms.

$$x=cx^{6}+5cx^{5}+6x^{4}c-4x^{3}c-8x^{2}c$$

Swap sides so that all variable terms are on the left hand side.

$$cx^{6}+5cx^{5}+6x^{4}c-4x^{3}c-8x^{2}c=x$$

Combine all terms containing c.

$$\left(x^{6}+5x^{5}+6x^{4}-4x^{3}-8x^{2}\right)c=x$$

Divide both sides by x^{6}+5x^{5}+6x^{4}-4x^{3}-8x^{2}.

$$\frac{\left(x^{6}+5x^{5}+6x^{4}-4x^{3}-8x^{2}\right)c}{x^{6}+5x^{5}+6x^{4}-4x^{3}-8x^{2}}=\frac{x}{x^{6}+5x^{5}+6x^{4}-4x^{3}-8x^{2}}$$

Dividing by x^{6}+5x^{5}+6x^{4}-4x^{3}-8x^{2} undoes the multiplication by x^{6}+5x^{5}+6x^{4}-4x^{3}-8x^{2}.

$$c=\frac{x}{x^{6}+5x^{5}+6x^{4}-4x^{3}-8x^{2}}$$

Divide x by x^{6}+5x^{5}+6x^{4}-4x^{3}-8x^{2}.

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