Lời giải:
\(a+b=1\Rightarrow a-1=-b; b-1=-a\)
Ta có:
\(P=\frac{a}{b^3-1}-\frac{b}{a^3-1}+\frac{2(a-b)}{a^2b^2+3}=\frac{a}{(b-1)(b^2+b+1)}-\frac{b}{(a-1)(a^2+a+1)}+\frac{2(a-b)}{a^2b^2+3}\)
\(=\frac{a}{-a(b^2+b+1)}-\frac{b}{-b(a^2+a+1)}+\frac{2(a-b)}{a^2b^2+3}\)
\(=\frac{1}{a^2+a+1}-\frac{1}{b^2+b+1}+\frac{2(a-b)}{a^2b^2+3}\)
\(=\frac{b^2-a^2+b-a}{(a^2+a+1)(b^2+b+1)}+\frac{2(a-b)}{a^2b^2+3}\)
\(=\frac{(b-a)(b+a+1)}{a^2b^2+ab(a+b)+a^2+b^2+a+b+ab+1}+\frac{2(a-b)}{a^2b^2+3}\)
\(=\frac{(b-a)(1+1)}{a^2b^2+ab+a^2+b^2+1+ab+1}+\frac{2(a-b)}{a^2b^2+3}\) (thay $a+b=1$)
\(=\frac{2(b-a)}{a^2b^2+(a+b)^2+2}+\frac{2(a-b)}{a^2b^2+3}=\frac{2(b-a)}{a^2b^2+1+2}+\frac{2(a-b)}{a^2b^2+3}\)
\(=\frac{2(b-a)}{a^2b^2+3}+\frac{2(a-b)}{a^2b^2+3}=0\)