Multiple radicals: $(\sqrt{10}+\sqrt{11}+\sqrt{12})(\sqrt{10}+\sqrt{11}-\sqrt{12})(\sqrt{10}-\sqrt{11}+\sqrt{12})(-\sqrt{10}+\sqrt{11}+\sqrt{12})$ $(\sqrt{10}+\sqrt{11}+\sqrt{12})(\sqrt{10}+\sqrt{11}-\sqrt{12})(\sqrt{10}-\sqrt{11}+\sqrt{12})(-\sqrt{10}+\sqrt{11}+\sqrt{12})$
I don't think multiplying these out will work, and I am stuck in the beginning, without a basic concept to get started. Can anyone show me how to do these? I would appreciate more detailed responses as I have a solution to this already but since it does not explain the steps, I cannot learn from it. Thanks



Answer, if you want to check

 359

 A: HINT:
$$(a+b+c)(a+b-c)(b+c-a)(c+a-b)$$
$$=\{(a+b)^2-c^2\}\{c^2-(a-b)^2\}$$
$$=-(a^2+b^2+2ab-c^2)(a^2+b^2-2ab-c^2)$$
$$=-\{(a^2+b^2-c^2)^2-(2ab)^2\}$$
which is symmetric on expansion.
So, WLOG choose $\{a,b,c\}$ from $\{\sqrt{10},\sqrt{11},\sqrt{12}\}$ to find the same result.
A: You can pair up the first and last items:
$$
(\sqrt{10}+\sqrt{11}+\sqrt{12})(-\sqrt{10}+\sqrt{11}+\sqrt{12})=
(\sqrt{11}+\sqrt{12})^2-10
$$
and the remaining two items
$$
(\sqrt{10}+\sqrt{11}-\sqrt{12})(\sqrt{10}-\sqrt{11}+\sqrt{12})=
10-(\sqrt{11}-\sqrt{12})^2
$$
The top expression becomes
$$
13+2\sqrt{11\cdot12}
$$
and the bottom one becomes
$$
-13+2\sqrt{11\cdot12}
$$
so the product is
$$
4\cdot11\cdot12-13^2
$$
A: It will be easier if you consider the expression 
\begin{align}
(a+b+c)(a+b-c)(a-b+c)(-a+b+c)&=[(a+b)^2-c^2][-(a-b)^2+c^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\\
&=4a^2b^2-a^4-b^4-c^4-2a^2b^2+2a^2c^2+2b^2c^2\\
&=2(a^2b^2+b^2c^2+c^2a^2)-(a^4+b^4+c^4)
\end{align}
And then you put $a=\sqrt{10}$, $b=\sqrt{11}$ and $c=\sqrt{12}$.
