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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 $c=29$, $d=-3$.

Jessie
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Liam Doan
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    Seems to me the only thing you can really nail down is $c-d$, not $c$ and $d$ separately (if you have one solution that works, then take that solution and increase both $c$ and $d$ with, say, $43$, that gives you another solution that works). Are you certain you copied it correctly? – Arthur Feb 28 '21 at 23:09
  • I'm certain - this is all the information provided. Is it possible that $c$ and $d$ could be any value as long as they satisfy $c-d=32$? – Liam Doan Feb 28 '21 at 23:10
  • I haven't looked that hard into the specifics of the problem, so I won't say for certain that $32$ is the right number. But other than that, yes, that's what it seems like to me. – Arthur Feb 28 '21 at 23:11
  • Please make your titles informative: they should be about the mathematical content of the post, not about what you want or are asking for personally. – Arturo Magidin Feb 28 '21 at 23:36
  • What has the title got to do with the question??? Does Q represent the rational numbers??? – MasB Mar 01 '21 at 00:15
  • What does $\equiv$ mean in this case. I get that $c= d+32$. But any $d$ will do. I imagine there must be somthing about $\equiv$ that will imply that $c=29$ and $d=-3$ are only options. (Although I don't see why) – fleablood Mar 01 '21 at 01:25

3 Answers3

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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.

jjagmath
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$4x^3 - 24x^2+48x + d = 4(x^3 - 6x^2 + 12x - 8) + 32+ d = 4(x-2)^3 + 32+ d\implies a = 4, b = -2, c = 32+d$. 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 !

  • When $x=1$ or $x=2$ it results in $d-c=32$. Do the values of $c$ and $d$ not matter, rather that they satisfy $d-c=-32$? – Liam Doan Feb 28 '21 at 23:18
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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.

  1. $a = 4$
  2. $3ab=-24$
  3. $3ab^2 = 48$.
  4. $ab^3 + c = d$.

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.

fleablood
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