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 ?

  • $\begingroup$ Which contest's question did your friend get it from?Is it from the BdMO 2006 Nationals? $\endgroup$
    – rah4927
    Aug 11, 2013 at 16:46
  • $\begingroup$ @rahul I don't know but it probably isn't from bdMO . $\endgroup$
    – A Googler
    Aug 11, 2013 at 16:57
  • $\begingroup$ Hmm. . .But I am sure this was the 6th question in the Junior section of the Bdmo 2006 nationals. $\endgroup$
    – rah4927
    Aug 11, 2013 at 17:03
  • $\begingroup$ @rahul really ? then it may be. Can you share the link to it ? $\endgroup$
    – A Googler
    Aug 11, 2013 at 17:32
  • $\begingroup$ You can find the question from BdMO's official website in the Questions section. $\endgroup$
    – rah4927
    Aug 11, 2013 at 17:36

5 Answers 5


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 $104$.

  • $\begingroup$ Delightful! ${}$ $\endgroup$ Aug 12, 2013 at 6:47
  • $\begingroup$ The poster wanted a solution without needing a calculator or using complex algebra, and the expression looked like the one in Heron's formula. $\endgroup$ Aug 12, 2013 at 12:51
  • 2
    $\begingroup$ Wow. I'm speechless. $\endgroup$
    – A Googler
    Aug 12, 2013 at 16:47

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 $$ (2\sqrt{42})^2-8^2=168-64=104. $$


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


for some coefficient $r$. It's easy to determine $r$ by letting $a=b=c=1$, for which we have


so $r=2$. The rest of the answer follows what the OP did:


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


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.)

  • $\begingroup$ Sorry , but I don't understand why P has no terms with odd degree . Can you please explain it ? $\endgroup$
    – A Googler
    Aug 11, 2013 at 17:35
  • $\begingroup$ @AGoogler, $P(a,b,c)=(P(a,b,c)+P(-a,b,c))/2$ rules out the possibility for $P$ to have any terms with $a$ of odd degree, and similarly for $b$ and $c$. $\endgroup$ Aug 11, 2013 at 19:06

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 $$=(2\sqrt{42}-8)(2\sqrt{42}+8)=(2\sqrt{42})^2-8^2=4\cdot42-64=104$$


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:

$$2(AB+AC+BC) - (A^2+B^2+C^2) = 2(30+35+42) - (25+36+49) = 214-110 = 104$$

  • 1
    $\begingroup$ What a great post for math! Many different ways to solve the same problem....everyone gets a plus 1!!!! $\endgroup$ Aug 12, 2013 at 2:56

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