Given y=x3...(i)
⇒dxdy=3x2
(dxdy)(−1,−1)=3
Equation of tangent at (–1,–1)
(y+1)=3(x+1)
y=3x+2…(ii)
Solving (i) and (ii)
⇒x3=3x+2
⇒x3−3x+2=0
⇒x=−1,−1,2
Another point of intersection is Q(2,8)
So, now plotting the diagram we get,

Required area from the above diagram will be,
=∫−12(3x+2−x3)dx
=23(4−1)+2(2+1)−41(16−1)
=427
Hence this is the required option.