Solve hx^3-x^4+3x^3+3x^2-x+1=0

$$h x ^ { 3 } - x ^ { 4 } + 3 x ^ { 3 } + 3 x ^ { 2 } - x + 1 = 0$$

Solve for h

$$h=x-3-\frac{3}{x}+\frac{1}{x^{2}}-\frac{1}{x^{3}}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Add x^{4} to both sides. Anything plus zero gives itself.

$$hx^{3}+3x^{3}+3x^{2}-x+1=x^{4}$$

Subtract 3x^{3} from both sides.

$$hx^{3}+3x^{2}-x+1=x^{4}-3x^{3}$$

Subtract 3x^{2} from both sides.

$$hx^{3}-x+1=x^{4}-3x^{3}-3x^{2}$$

Add x to both sides.

$$hx^{3}+1=x^{4}-3x^{3}-3x^{2}+x$$

Subtract 1 from both sides.

$$hx^{3}=x^{4}-3x^{3}-3x^{2}+x-1$$

The equation is in standard form.

$$x^{3}h=x^{4}-3x^{3}-3x^{2}+x-1$$

Divide both sides by x^{3}.

$$\frac{x^{3}h}{x^{3}}=\frac{x^{4}-3x^{3}-3x^{2}+x-1}{x^{3}}$$

Dividing by x^{3} undoes the multiplication by x^{3}.

$$h=\frac{x^{4}-3x^{3}-3x^{2}+x-1}{x^{3}}$$

Divide x^{4}-3x^{3}-3x^{2}+x-1 by x^{3}.

$$h=x-3+\frac{-3x^{2}+x-1}{x^{3}}$$