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 ?

As $$(a+b+c)(-a+b+c)=\{(b+c)+a\}\{(b+c)-a\}=(b+c)^2-a^2,$$
$$(\sqrt5 +\sqrt6 +\sqrt7)(−\sqrt5+\sqrt6+\sqrt7)$$
$$=(\sqrt6+\sqrt7)^2-(\sqrt5)^2=6+7+2\sqrt7\cdot\sqrt6-5=8+2\sqrt{42}=2\sqrt{42}+8$$
Again as  $$(a-b+c)(a+b-c)=\{a+(b-c)\}\{a-(b-c)\}=a^2-(b-c)^2,$$
$$(\sqrt5 −  \sqrt6 + \sqrt7)(\sqrt5 + \sqrt6 − \sqrt7)=\{\sqrt5 −  (\sqrt6-\sqrt7)\}\{\sqrt5 + (\sqrt6-\sqrt7)\}$$
$$=(\sqrt5)^2-(\sqrt6 − \sqrt7)^2=5-(6+7-\sqrt7\cdot\sqrt6)=2\sqrt{42}-8$$
So, the product $$=(2\sqrt{42}-8)(2\sqrt{42}+8)=(2\sqrt{42})^2-8^2=4\cdot42-64=104$$
