<s>[INST] The value of $(a^2+b^2)(c^2+1)$. The question is:

Given real numbers $a,b,c$ that satisfy
$$ab(c^2-1)+c(a^2-b^2)=12$$
$$(a+b)c+(a-b)=7$$
Find the value of $(a^2+b^2)(c^2+1)$

From what I've done, I got $7(3ac+3a+3bc-b)-2ab(c+1)(c-1)=(a^2+b^2)(c^2+1)$. I think I'm inching further from the actual solution. Can anyone give me a hint? Thanks in advance :).
 [/INST] Since you are asking for the value of $(a^2+b^2)(c^2+1)$, we may assume that the value is equal for all $b$, so we can just set $b=0$, so that $(a-3)(a-4)=0$ and $c=12/a^2$. For $a=3$ we obtain $c=4/3$, so that $(a^2+b^2)(c^2+1)=16+9=25$. The other case, $a=4$ gives $c=3/4$ and again $(a^2+b^2)(c^2+1)=9+16=25$.
But even if we do not assume that there is only one value, we can generally conclude that one has either $a=3-bc$ or $a=4-bc$, so that again $(a^2+b^2)(c^2+1)=25$.
So the answer is $25$.
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