\(\frac{a}{b³-1}\) - \(\frac{b}{a³-1}\)
= \(\frac{1-b}{b³-1}\) - \(\frac{1-a}{a³-1}\)
= \(\frac{1-b}{(b-1)(b²+b+1)}\) - \(\frac{1-a}{(a-1)(a²+a+1)}\)
=\(\frac{a-1}{(a-1)(a²+a+1)}\)- \(\frac{b-1}{(b-1)(b²+b+1)}\)
= \(\frac{1}{a²+a+1}\) - \(\frac{1}{b²+b+1}\)
= \(\frac{b²+ b +1- a² -a-1}{a²b²+ab(a+b)+(a+b)+a²+b²+ab+1}\)
= \(\frac{(b-a)(b+a)+(b-a)}{a²b²+ 2ab+a² + b² +2}\)
= \(\frac{(a+b+1)(b-a)}{a²b² + 2 + (a+b)²}\)
= \(\frac{2(b-a)}{a²b² + 3}\) (đpcm)
