Question

Express the following algebraic fractions in the reduced from:

$$\frac{a^2-ab+b^2}{a^2+ab}÷\frac{a^3+b^3}{a^2-b^2}$$

Answer 1

$$\frac{a^2-ab+b^2}{a^2+ab}÷\frac{a^3+b^3}{a^2-b^2}=\frac{a^2-ab+b^2}{a^2+ab}\times \frac{a^2-b^2}{a^3+b^3}$$
$$\begin{array}{cc}& =\frac{(a^2-ab+b^2)(a^2-b^2)}{(a^2+ab)(a^3+b^3)}\\ & =\frac{(a^2-ab+b^2)(a+b)(a-b)}{\left(\right(a\left)\right(a+b\left)\right)\left(\right(a+b\left)(a^2-ab+b^2)\right)}\end{array}$$
Cancelling same terms from numerator and denominator, we get
$$=\frac{(a-b)}{a(a+b)}$$
Therefore, $\frac{a^2-ab+b^2}{a^2+ab}÷\frac{a^3+b^3}{a^2-b^2}=\frac{(a-b)}{a(a+b)}$