Question

$(\frac{a ^{2} -2ab + b ^{2} } {a^{3}+ b ^{3} } - \frac{a-b} {a ^{2} -ab + b ^{2} }} ) : \frac{2ab}{a ^{2} -ab + b ^{2} }}$
Прошу помогите

Answer 1

$(\frac{a ^{2} -2ab + b ^{2} } {a^{3}+ b ^{3} } - \frac{a-b} {a ^{2} -ab + b ^{2} }} ) : \frac{2ab}{a ^{2} -ab + b ^{2} }}= \\ \\ =(\frac{a ^{2} -2ab + b ^{2} } {(a+b)(a^{2}-ab+ b ^{2}) } - \frac{(a-b)(a+b)} {(a+b)(a ^{2} -ab + b ^{2} )}} ) \cdot \frac{a ^{2} -ab + b ^{2} }{2ab}=$

$=(\frac{a ^{2} -2ab + b ^{2} -a ^{2}+b ^{2} } {(a+b)(a^{2}-ab+ b ^{2}) } ) \cdot \frac{a ^{2} -ab + b ^{2} }{2ab}=(\frac{2b ^{2} -2ab } {(a+b) } ) \cdot \frac{1 }{2ab}= \\ \\ =(\frac{2b(b-a) } {(a+b) } ) \cdot \frac{1 }{2ab}= \frac{b-a}{a(a+b)}$

Answer 2

$(\frac{a^2 -2ab+b^2}{(a+b) (a^2-ab+b^2)} - \frac{a-b}{a^2-ab+b^2}) \cdot \frac{a^2-ab+b^2}{2ab}=\frac{a^2 -2ab+b^2- a^2 -b^2}{(a+b) (a^2 -ab+b^2)} \cdot \frac{a^2-ab+b^2}{2ab}= \\ \\ =- \frac{1}{a+b}$