Simplify : $( \sqrt 5 + \sqrt6 + \sqrt7)(− \sqrt5 + \sqrt6 + \sqrt7)(\sqrt5 − \sqrt6 + \sqrt7)(\sqrt5 + \sqrt6 − \sqrt7) $ The question is to simplify  $(  \sqrt 5 + \sqrt6 + \sqrt7)(−  \sqrt5 + \sqrt6 + \sqrt7)(\sqrt5 −  \sqrt6 + \sqrt7)(\sqrt5 + \sqrt6 − \sqrt7)$ without using a calculator .
  My friend has given me this challenge . I solved it by expanding $$(a+b+c)(-a+b+c)(a-b+c)(a+b-c) = -a^4+2a^2b^2+2a^2c^2-b^4+2b^2c^2-c^4$$ and then substituting $a,b,c=\sqrt 5 , \sqrt6 , \sqrt7$ respectively to get the answer $104$ .
But I suppose there is a more elegant and easy way to solve this problem .
Can anyone find it ?

Consider Heron's formula:  the area of a triangle with sides $a, b, \text{and } c$ is
$$
\sqrt{s(s-a)(s-b)(s-c)}
$$
where $s$ is the semi-perimeter $\frac12 (a + b + c)$.
Let $a, b, \text{and } c$ be $\sqrt{5}, \sqrt{6}, \text{and } \sqrt{7}$. Then the area is the square root of your expression divided by $4$.  So, what is the area of this triangle?  Use the law of cosines to find the cosine of the angle $C$ opposite $c$:
$$
\begin{align}
7 &= 5 + 6 - 2 \sqrt{5}\sqrt{6}\cos{C}\\
2\sqrt{30}\cos{C} &= 4\\
\cos{C} &= \frac{2}{\sqrt{30}}
\end{align}
$$
But the area of the triangle is $\frac12 ab\sin{C}$.
$$
\frac12 ab\sin{C} = \frac12 \sqrt{30} \frac{\sqrt{26}}{\sqrt{30}} = \frac12\sqrt{26}.
$$
Your expression is therefore the square of $2\sqrt{26}$, which is $104$.
