Let f be an invertible function from R→R satisfying the equation f3x−x3+2f2x+2x3+1fx−x3=0
Then the value of f'(8)×f−1'(8), is :
12
16
20
32
(f(x)−1)2f(x)−x3=0f(x)=x3f′(x)=3x2