Solve x^3/(x^2+4)^2=Ax+B/(x^2+4)

$$\frac { x ^ { 3 } } { ( x ^ { 2 } + 4 ) ^ { 2 } } = \frac { A x + B } { ( x ^ { 2 } + 4 ) }$$

Solve for A

$$\left\{\begin{matrix}A=-\frac{4B+Bx^{2}-x^{3}}{x\left(x^{2}+4\right)}\text{, }&x\neq 0\\A\in \mathrm{R}\text{, }&x=0\text{ and }B=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply both sides of the equation by \left(x^{2}+4\right)^{2}, the least common multiple of \left(x^{2}+4\right)^{2},x^{2}+4.

$$x^{3}=\left(x^{2}+4\right)\left(Ax+B\right)$$

Use the distributive property to multiply x^{2}+4 by Ax+B.

$$x^{3}=Ax^{3}+x^{2}B+4Ax+4B$$

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

$$Ax^{3}+x^{2}B+4Ax+4B=x^{3}$$

Subtract x^{2}B from both sides.

$$Ax^{3}+4Ax+4B=x^{3}-x^{2}B$$

Subtract 4B from both sides.

$$Ax^{3}+4Ax=x^{3}-x^{2}B-4B$$

Reorder the terms.

$$Ax^{3}+4Ax=x^{3}-Bx^{2}-4B$$

Combine all terms containing A.

$$\left(x^{3}+4x\right)A=x^{3}-Bx^{2}-4B$$

Divide both sides by x^{3}+4x.

$$\frac{\left(x^{3}+4x\right)A}{x^{3}+4x}=\frac{x^{3}-Bx^{2}-4B}{x^{3}+4x}$$

Dividing by x^{3}+4x undoes the multiplication by x^{3}+4x.

$$A=\frac{x^{3}-Bx^{2}-4B}{x^{3}+4x}$$

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

$$A=\frac{x^{3}-Bx^{2}-4B}{x\left(x^{2}+4\right)}$$

Solve for B

$$B=-\frac{x\left(Ax^{2}-x^{2}+4A\right)}{x^{2}+4}$$

Steps for Solving Linear Equation

Multiply both sides of the equation by \left(x^{2}+4\right)^{2}, the least common multiple of \left(x^{2}+4\right)^{2},x^{2}+4.

$$x^{3}=\left(x^{2}+4\right)\left(Ax+B\right)$$

Use the distributive property to multiply x^{2}+4 by Ax+B.

$$x^{3}=Ax^{3}+x^{2}B+4Ax+4B$$

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

$$Ax^{3}+x^{2}B+4Ax+4B=x^{3}$$

Subtract Ax^{3} from both sides.

$$x^{2}B+4Ax+4B=x^{3}-Ax^{3}$$

Subtract 4Ax from both sides.

$$x^{2}B+4B=x^{3}-Ax^{3}-4Ax$$

Reorder the terms.

$$Bx^{2}+4B=-Ax^{3}+x^{3}-4Ax$$

Combine all terms containing B.

$$\left(x^{2}+4\right)B=-Ax^{3}+x^{3}-4Ax$$

Divide both sides by x^{2}+4.

$$\frac{\left(x^{2}+4\right)B}{x^{2}+4}=\frac{x\left(-Ax^{2}+x^{2}-4A\right)}{x^{2}+4}$$

Dividing by x^{2}+4 undoes the multiplication by x^{2}+4.

$$B=\frac{x\left(-Ax^{2}+x^{2}-4A\right)}{x^{2}+4}$$