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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

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

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HINT:

If $F(t)=c_0+c_1t+c_2t^2+\cdots$ then $$ c_n=\frac1{n!}\left.\frac{d^nF}{dt^n}\right|_{t=0}. $$

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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 $$ ax^3 + (3a + b)x^2 + (3a + 2b + c)x + a + b + c + d. $$

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