Evaluate $(\sqrt{5}+\sqrt{6}+\sqrt{7})(\sqrt{5}+\sqrt{6}-\sqrt{7})(\sqrt{5}-\sqrt{6}+\sqrt{7})(-\sqrt{5}+\sqrt{6}+\sqrt{7})$ Evaluate $(\sqrt{5}+\sqrt{6}+\sqrt{7})(\sqrt{5}+\sqrt{6}-\sqrt{7})(\sqrt{5}-\sqrt{6}+\sqrt{7})(-\sqrt{5}+\sqrt{6}+\sqrt{7})$
I find out the answer $104$ but my friend find the answer $96$ using calculator.which one is correct?
 A: Generally:
$$\begin{aligned}
&\phantom{mm}(a+b+c)(a+b-c)(a-b+c)(-a+b+c)\\
&=((a+b)+c)((a+b)-c) \times ( c+(a-b))(c-(a-b)) \\
&=((a+b)^2 -c^2)(c^2-(a-b)^2)\\
&=[2ab+(a^2+b^2 - c^2) ] [ 2ab - (a^2+b^2-c^2)]\\
&= 4a^2b^2 - (a^2+b^2-c^2)^2.
\end{aligned}$$
Here $a^2=5, b^2=6,$ and $c^2=7$, so the answer is $104$ as you said.
A: By Heron's formula, the quantity equals $16$ times the square of the area of a triangle of sides $\surd5,\surd6$, and $\surd7$. You can evaluate this also by the "half base times height" formula, where the squared height is easily calculated, by Pythagoras, to be $26/7$ from the $\surd7$ base. This confirms the result $104$.
A: Let $x=\sqrt7$ and $y=\sqrt5+\sqrt6$ and $z=\sqrt5-\sqrt6$. It becomes:
$$(y+x)(y-x)(x+z)(x-z)$$
$$=(y^2-x^2)(x^2-z^2)$$
$$=-x^4+x^2(y^2+z^2)-y^2z^2$$
$$=-7^2+7((\sqrt5+\sqrt6)^2+(\sqrt5-\sqrt6)^2)-(\sqrt5+\sqrt6)^2(\sqrt5-\sqrt6)^2$$
$$=-49+7(5+2\sqrt5\sqrt6+6+5-2\sqrt5\sqrt6+6)-(5-6)^2$$
$$=-49+7\times22-1$$
$$=104$$
A: Hint:
let $x=\sqrt(5)+\sqrt(6)$ and $y=\sqrt(5)-\sqrt(6)$
so 
$$(x+\sqrt{7})(x-\sqrt{7})(y+\sqrt{7})(-y+\sqrt{7})$$
then you can use the fact of
$$(a-b)(a+b)=a^2-b^2$$
