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Aus Online Mathematik Brückenkurs 2

If we focus on the leading term x3, we need to complement it with ax2 in order to get a sub expression that is divisible by the denominator,

x+ax3+a3=x+ax3+ax2−ax2+a3.

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 −ax2+a3 in the numerator,

x+ax3+ax2−ax2+a3=x+ax3+ax2+x+a−ax2+a3=x+ax2(x+a)+x+a−ax2+a3=x2+x+a−ax2+a3.

When we treat the new quotient, we add and take away −a2x to/from −ax2 in order to get something divisible by x+a,

x2+x+a−ax2+a3=x2+x+a−ax2−a2x+a2x+a3=x2+x+a−ax2−a2x+x+aa2x+a3=x2+x+a−ax(x+a)+x+aa2x+a3=x2−ax+x+aa2x+a3.

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

x2−ax+x+aa2x+a3=x2−ax+x+aa2(x+a)=x2−ax+a2.

If we have calculated correctly, we should have

x+ax3+a3=x2−ax+a2

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

x3+a3=(x2−ax+a2)(x+a).

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

RHS=(x2−ax+a2)(x+a)=x3+ax2−ax2−a2x+a2x+a3=x3+a3=LHS.