f(x) = (x+1)3 (x-1)3
f'(x) = 3(x+1)2 (x-1)3 + 3(x+3)3(x-1)2
= 3(x+1)2(x-1)2(x-1+x+1)
= 6x(x+1)2(x-1)2
∵ f'(x) > 0 when x > 0 & x ≠ 1
& f'(x) < 0 when x < 0 & x ≠ -1
Hence,
f(x) is strictly increasing in (0,∞) - {1}
& stictly decreasing in (-∞,0) - {-1}