# Given $(x-1)^3+3(x-1)^2-2(x-1)-4=a(x+1)^3+b(x+1)^2+c(x+1)+d$, find$(a,b,c,d)$

Given $$(x-1)^3+3(x-1)^2-2(x-1)-4=a(x+1)^3+b(x+1)^2+c(x+1)+d$$, find$$(a,b,c,d)$$

my attempt: $$(x+1)=(x-1)\frac{(x+1)}{(x-1)}$$ but this seems useless? I want to use synthetic division but I don't know how

• You could simply substitute $x=y-1$ and expand the LHS. – Martin R Jul 24 at 11:46
• The LHS simplifies to $x^3 - 5x$. – Viktor Glombik Jul 24 at 11:48

It's $$(x+1-2)^3+3(x+1-2)^2-2(x+1-2)-4=(x+1)^3-3(x+1)^2-2(x+1)+4.$$ Can you end it now?
If $$F(t)=c_0+c_1t+c_2t^2+\cdots$$ then $$c_n=\frac1{n!}\left.\frac{d^nF}{dt^n}\right|_{t=0}.$$
In case you don't know how to solve this most elegantly, there is still the straightforward possibility by expanding both sides. The LHS is given by $$x^3-5x,$$ whereas the RSH is $$ax^3 + (3a + b)x^2 + (3a + 2b + c)x + a + b + c + d.$$