(5.) \( \frac{a^{3}+b^{3}+a b^{2}-3 a^{2} b}{a^{2}-2 a b+b^{2}}= \)
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To simplify the expression \( \frac{a^{3}+b^{3}+a b^{2}-3 a^{2} b}{a^{2}-2 a b+b^{2}} \), we start by factoring the numerator and the denominator. The denominator, \( a^{2} - 2ab + b^{2} \), can be recognized as a perfect square. It factors to \( (a-b)^{2} \). For the numerator, we can rearrange it and factor by grouping: \[ a^{3} - 3a^{2}b + b^{3} + ab^{2} = (a^{3} - 3a^{2}b + 3ab^{2} - b^{3}) + (b^{3} + ab^{2} - 3ab^{2}) \] Notice that we can combine the terms strategically. An effective way to factor \( a^{3} + b^{3} \) is to employ the identity \( a^{3} + b^{3} = (a+b)(a^{2}-ab+b^{2}) \), and combine it with the remaining terms. So, let’s write it clearly: \[ = (a+b)(a^{2}-ab+b^{2}) + ab^{2} - 3a^{2}b \] The whole expression simplifies via polynomial long division or more straightforward factorizable patterns, yielding: \[ = \frac{(a-b)(a^2 + ab + b^2)}{(a-b)^2} \] Upon simplifying, we obtain: \[ a^2 + ab + b^2 = \frac{a^2 + ab + b^2}{(a-b)} \cdots (a \neq b) \] Thus the overall result is: \[ = \frac{a^2 + ab + b^2}{(a-b)} \] Remembering that the values of \( a \) and \( b \) must be chosen carefully as their values allow the factorization without resulting in division by zero.
