Solve 3x-A(A^3/A+A^2)=9-A^2

$$3 x - A ( \frac { A ^ { 3 } } { A + A ^ { 2 } } ) = 9 - A ^ { 2 }$$

Solve for x

$$x=-\frac{A^{2}-9A-9}{3\left(A+1\right)}$$

$$A\neq -1\text{ and }A\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by A\left(A+1\right).

$$3xA\left(A+1\right)-AA^{3}=A\left(A+1\right)\times 9-A^{2}A\left(A+1\right)$$

To multiply powers of the same base, add their exponents. Add 1 and 3 to get 4.

$$3xA\left(A+1\right)-A^{4}=A\left(A+1\right)\times 9-A^{2}A\left(A+1\right)$$

Use the distributive property to multiply 3xA by A+1.

$$3xA^{2}+3xA-A^{4}=A\left(A+1\right)\times 9-A^{2}A\left(A+1\right)$$

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$3xA^{2}+3xA-A^{4}=A\left(A+1\right)\times 9-A^{3}\left(A+1\right)$$

Use the distributive property to multiply A by A+1.

$$3xA^{2}+3xA-A^{4}=\left(A^{2}+A\right)\times 9-A^{3}\left(A+1\right)$$

Use the distributive property to multiply A^{2}+A by 9.

$$3xA^{2}+3xA-A^{4}=9A^{2}+9A-A^{3}\left(A+1\right)$$

Use the distributive property to multiply -A^{3} by A+1.

$$3xA^{2}+3xA-A^{4}=9A^{2}+9A-A^{4}-A^{3}$$

Add A^{4} to both sides.

$$3xA^{2}+3xA=9A^{2}+9A-A^{4}-A^{3}+A^{4}$$

Combine -A^{4} and A^{4} to get 0.

$$3xA^{2}+3xA=9A^{2}+9A-A^{3}$$

Combine all terms containing x.

$$\left(3A^{2}+3A\right)x=9A^{2}+9A-A^{3}$$

The equation is in standard form.

$$\left(3A^{2}+3A\right)x=9A+9A^{2}-A^{3}$$

Divide both sides by 3A^{2}+3A.

$$\frac{\left(3A^{2}+3A\right)x}{3A^{2}+3A}=\frac{A\left(9+9A-A^{2}\right)}{3A^{2}+3A}$$

Dividing by 3A^{2}+3A undoes the multiplication by 3A^{2}+3A.

$$x=\frac{A\left(9+9A-A^{2}\right)}{3A^{2}+3A}$$

Divide A\left(9A+9-A^{2}\right) by 3A^{2}+3A.

$$x=\frac{9+9A-A^{2}}{3\left(A+1\right)}$$

Solve for A (complex solution)

$$A=\frac{\sqrt{9x^{2}-66x+117}-3x+9}{2}$$

$$A=\frac{-\sqrt{9x^{2}-66x+117}-3x+9}{2}\text{, }x\neq 3$$

Solve for A

$$A=\frac{\sqrt{9x^{2}-66x+117}-3x+9}{2}$$

$$A=\frac{-\sqrt{9x^{2}-66x+117}-3x+9}{2}\text{, }x\geq \frac{13}{3}\text{ or }x<3$$