Aus Online Mathematik Brückenkurs 2

If we focus on the leading term \displaystyle x^{3}, we need to complement it with in order to get an expression that is divisible by the denominator ,

\displaystyle \frac{x^{3}+a^{3}}{x+a}=\frac{x^{3}+ax^{2}-ax^{2}+a^{3}}{x+a}

With this form on the right-hand side, we can separate away the first two terms in the numerator and have left a polynomial quotient with \displaystyle -ax^{2}+a^{3} in the numerator:

\displaystyle \begin{align} & \frac{x^{3}+ax^{2}-ax^{2}+a^{3}}{x+a}=\frac{x^{3}+ax^{2}}{x+a}+\frac{-ax^{2}+a^{3}}{x+a} \\ & =\frac{x^{2}\left( x+a \right)}{x+a}+\frac{-ax^{2}+a^{3}}{x+a} \\ & =x^{2}+\frac{-ax^{2}+a^{3}}{x+a} \\ \end{align}

When we treat the new quotient, we add and take away \displaystyle -a^{2}x to/from \displaystyle -ax^{2} in order to get something divisible by \displaystyle x+a

\displaystyle \begin{align} & x^{2}+\frac{-ax^{2}+a^{3}}{x+a}=x^{2}+\frac{-ax^{2}-a^{2}x+a^{2}x+a^{3}}{x+a} \\ & =x^{2}+\frac{-ax^{2}-a^{2}x}{x+a}+\frac{a^{2}x+a^{3}}{x+a} \\ & =x^{2}+\frac{-ax\left( x+a \right)}{x+a}+\frac{a^{2}x+a^{3}}{x+a} \\ & =x^{2}-ax+\frac{a^{2}x+a^{3}}{x+a} \\ \end{align}

In the last quotient, the numerator has \displaystyle x+a as a factor, and we obtain a perfect division:

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

If we have calculated correctly, we should have

\displaystyle \frac{x^{3}+a^{3}}{x+a}=x^{2}-ax+a^{2}

and one way to check the answer is to multiply both sides by \displaystyle x+a,

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

Then, expand the right-hand side and then we should get what is on the left-hand side:

\displaystyle \begin{align} & \text{RHS}=\left( x^{2}-ax+a^{2} \right)\left( x+a \right)=x^{3}+ax^{2}-ax^{2}-a^{2}x+a^{2}x+a^{3} \\ & =x^{3}+a^{3}=~~\text{LHS} \\ \end{align}