Solve the cubic equation $\displaystyle 3x^3-x^2-8x-4=0$. Hence solve the equation $\displaystyle 3(y+1)^3-(y+1)^2=8y+12$

I have already solved the cubic equation however the 2nd part makes no sense to me.

Answers to cubic equation is $\displaystyle (x+1)(3x+2)(x-2)$

MY GUESS is that the solution to solving part 2 is changing the y in 8y+12 to (y+1) so as to substitute y+1 = x