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 :).

We have
$$
(a+b)^2c^2+2(a^2-b^2)c + (a-b)^2=49
\\2ab(c^2-1) +2(a^2-b^2)c = 24
$$
Subtract them and we get
$$(a^2+b^2)c^2+2ab+(a-b)^2=25$$
Thus $$(a^2+b^2)(c^2+1) = 25.$$
