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Aus Online Mathematik Brückenkurs 2

If we focus on the leading term x3, we need to complement it with in order to get an 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+ax2x+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−axx+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+aa2x+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+a2x+a 

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

RHS=x2−ax+a2x+a=x3+ax2−ax2−a2x+a2x+a3=x3+a3=  LHS