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 ?
 A: Given that your original numbers are square roots and your expanded equation involves only even powers of your original numbers, you could say:
$$A=a^2=5, B=b^2=6, C=c^2=7$$
and simplify your equation to:
$$2(AB+AC+BC) - (A^2+B^2+C^2) = 2(30+35+42) - (25+36+49) = 214-110 = 104$$
A: 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 $104$.
A: Use $$(a+b)(a-b)=a^2-b^2.$$
With $a=\sqrt6+\sqrt7$, $b=\sqrt5$ you see that the product of first two numbers is $(\sqrt6+\sqrt7)^2-5=8+2\sqrt{42}$. With $a=\sqrt5$, $b=\sqrt7-\sqrt6$ you get that
the product of last two is $5-(\sqrt7-\sqrt6)^2=-8+2\sqrt{42}$. One last application of this rule tells you that the answer is
$$
(2\sqrt{42})^2-8^2=168-64=104.
$$
A: Here's a nice way to get the expansion the OP used:
The expression 
$$P(a,b,c)=(a+b+c)(-a+b+c)(a-b+c)(a+b-c)$$
is clearly a homogeneous polynomial of degree $4$, symmetric in its three variables.  It's also clear that the coefficient of $a^4$ (hence also $b^4$ and $c^4$) is $-1$.  Moreover, 
$$P(-a,b,c)=P(a,-b,c)=P(a,b,-c)=P(a,b,c)$$ which implies $P$ has no terms with any variable taken to an odd degree.  Therefore $P$ must be of the form
$$P(a,b,c)=r(a^2b^2+b^2c^2+c^2a^2)-(a^4+b^4+c^4)$$
for some coefficient $r$.  It's easy to determine $r$ by letting $a=b=c=1$, for which we have 
$$3=3\cdot1\cdot1\cdot1=P(1,1,1)=r(1+1+1)-(1+1+1)=3r-3$$
so $r=2$.  The rest of the answer follows what the OP did:
$$P(\sqrt5,\sqrt6,\sqrt7)=2(5\cdot6+6\cdot7+7\cdot5)-(5^2+6^2+7^2)=104$$
Added 8/12/13:  Eric Jablow's invocation of Heron's formula inspires one more approach:
Consider the triangle formed by the origin and two vectors $x$ and $y$, with $|x|=a$, $|y|=b$, and $|x-y|=c$.  There are two formulas for the area of the triangle:  Heron's formula
$$\sqrt{s(s-a)(s-b)(s-c)}$$
where $s=(a+b+c)/2$, and the dot-product formula
$${1\over2}\sqrt{|x|^2|y|^2-(x\cdot y)^2}$$
Putting these together, we have
$$(a+b+c)(-a+b+c)(a-b+c)(a+b-c)=4((a^2b^2-(x\cdot y)^2)$$
What's nice is that this holds for vectors $x$ and $y$ in any dimension.  So now let $x=(2,1,0,0)$ and $y=(0,2,1,1)$, so that $x-y=(2,-1,-1,-1)$.  We have $a=\sqrt5$, $b=\sqrt6$, $c=\sqrt7$, and $x\cdot y=2$, and thus the OP's product of square roots simplifies to
$$4(5\cdot6-2^2) = 104$$
If you want to know why we went into the fourth dimension for the vectors $x$ and $y$, it's because $7$, like all positive integers, can be written as the sum of four squares, but not as the sum of three.  Another possibility, which makes it clear one can handle arbitrary $a$, $b$, and $c$ (as long as the triangle inequality is satisfied, at least) is to let $x=(1,1,1,1,1,0,0,0,0)$ and $y=(0,0,0,1,1,1,1,1,1)$, so that $x-y=(1,1,1,0,0,-1,-1,-1,-1)$. (Edit:  Actually, what I just said only works when the triangle inequality holds and $a+b+c$ is even.)
A: 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 $$=(2\sqrt{42}-8)(2\sqrt{42}+8)=(2\sqrt{42})^2-8^2=4\cdot42-64=104$$
