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 $$(a^2+b^2)(c^2+1) = 25.$$
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 $25$.
How can we ask these questions to WA? https://www.wolframalpha.com/input?i=%28%28a%2Bb%29c%2B%28a-b%29%29%5E2-2*%28%28ab%28c%5E2-1%29%2Bc%28a%5E2-b%5E2%29%29 Anyway. According to this, the answer is $7^2-2*12=25$.
