Question

1 простое задание по алгебре
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Answer 1

Замена $a=b^3$

$\frac{1}{2}\left(\frac{b^3}{b^3+8}-\frac{4b^3}{(b+2)^3}\right)\cdot \left(\frac{b+2}{b-2}\right)-\frac{12}{b^3+8}=$

$=\frac{b^3}{2}\cdot \left(\frac{1}{(b+2)(b^2-2b+4)}-\frac{4}{(b+2)^3}\right)\cdot \frac{b+2}{b-2}-\frac{12}{b^3+8}=$

$=\frac{b^3}{2}\cdot\frac{(b+2)^2-4(b^2-2b+4)}{(b^3+8)(b+2)^2}\cdot \frac{b+2}{b-2}-\frac{12}{b^3+8}=$

$=\frac{b^3}{2}\cdot\frac{b^2+4b+4-4b^2+8b-16}{(b^3+8)(b+2)(b-2)}-\frac{12}{b^3+8}=$

$=\frac{b^3(-3b^2+12b-12)}{2(b^3+8)(b+2)(b-2)}-\frac{12}{b^3+8}= \frac{-3b^3(b-2)^2}{2(b^3+8)(b+2)(b-2)}-\frac{12}{b^3+8}=$

$=\frac{-3b^3(b-2)-24(b+2)}{2(b^3+8)(b+2)}= \frac{-3(b^4-2b^3+8b+16)}{2(b^3+8)(b+2)}= \frac{-3(a^{4/3}-2a+8a^{1/3}+16)}{2(a+8)(a^{1/3}+2)}$