Expand the right hand side, simplify, and match up the coefficients of same-power terms.
[tex]a(x+1)^3+b(x+1)^2+c(x+1)=a(x^3+3x^2+3x+1)+b(x^2+2x+1)+c(x+1)[/tex]
[tex]=ax^3+(3a+b)x^2+(3a+2b+c)x+(a+b+c)[/tex]
Since this must be the same as [tex]2x^3+3x^2-4x-5[/tex], you need to satisfy
[tex]\begin{cases}a=2\\3a+b=3\\3a+2b+c=-4\\a+b+c=-5\end{cases}[/tex]
The solution for [tex]a[/tex] is already given, [tex]a=2[/tex]. Substitute this into the second equation to find [tex]b[/tex].
[tex]3a+b=3\implies 6+b=3\implies b=-3[/tex]
Now plug [tex]a,b[/tex] into either remaining equation to find [tex]c[/tex].
[tex]a+b+c=-5\implies 2-3+c=-5\implies c=-4[/tex]
And just to check that the third equation is also satisfied,
[tex]3a+2b+c=-4\implies 6-6-4=-4\checkmark[/tex]
