Determine the value of each of the constants $a$, $b$, $c$, and $d$ in the identity $a(x+b)^3+c\equiv4x^3-24x^2+48x+d$ 
Determine the value of each of the constants $a$, $b$, $c$, and $d$ in the identity
$$a(x+b)^3+c\equiv4x^3-24x^2+48x+d$$

I have already found $a=4$ and $b=-2$ but I'm struggling to find $c$ and $d$.
The answers for $c$ and $d$ are $c=29$, $d=-3$.
 A: Set $a=4$, $b=-2$, $c=32+t$ and $d=t$ and you'll get that the identity is true for all $t$ which means the solution is not unique, in other words, there's no enough information to get the supposed answer you were given.
A: $4x^3 - 24x^2+48x + d = 4(x^3 - 6x^2 + 12x - 8) + 32+ d = 4(x-2)^3 + 32+ d\implies a = 4, b = -2, c = 32+d$. Note that there are more than one right answer for this. You can take for example $c = 3, d = -29$ or $c = 0, d = -32$ as they are just two answers among many !
A: Expanding we get $ax^3 + 3abx^2 + 3ab^2x +ab^3 + c = 4x^2-24x^2 +48x + d$
Technically what you have going on is 4  equations.

*

*$a = 4$

*$3ab=-24$

*$3ab^2 = 48$.

*$ab^3 + c = d$.

They aren't linear as the involve $b$ to powers $2$ and $3$ but $1,2$ are independent linear so $a,b$ can be solved $a= 4, b=-2$. And a equation three is completely redundant (it simplifies to the inane and unhelpful $48 = 48$... gee, no ****)
That leaves a final equation $-32 + c =  d$.  This is the only equation with $c$ or $d$ so we can express $c$ in terms of $d$  ($c =d +32$) or $d$ in terms of $c$ ($d=c-32$) but as it is one equation with two unknowns that's as for as we can go. We can let $d$ be anything we want and $c$ infinite potential values will follow (or vice versa.  $c=29$ and $d = c -32 =-3$ is certainly an answer but it is by no means the only solution.
After all if we had replaced $c,d$ with $M,N$ or $K + 856, J+ 856$ or $c -5, d-5$ nothing would have changed.
