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$(\sqrt{10}+\sqrt{11}+\sqrt{12})(\sqrt{10}+\sqrt{11}-\sqrt{12})(\sqrt{10}-\sqrt{11}+\sqrt{12})(-\sqrt{10}+\sqrt{11}+\sqrt{12})$

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

Martin Sleziak
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suomynonA
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    For $(\sqrt{10}+\sqrt{11}+\sqrt{12})(\sqrt{10}+\sqrt{11}-\sqrt{12})$ use the formula for difference of two squares so it is $(\sqrt{10}+\sqrt{11})^2-(\sqrt{12})^2$, perhaps this is a start? The other two terms also form a difference of squares $(\sqrt{12}+\sqrt{10}-\sqrt{11})(\sqrt{12}-(\sqrt{10}-\sqrt{11}))$. – Mirko Mar 11 '16 at 16:17
  • "I don't think multiplying these out will work": try a minimum, there remarkable products... –  Mar 11 '16 at 16:31
  • Mentally by exploiting remarkable products, $(9+2\sqrt{110})(-9+2\sqrt{110})=359$. –  Mar 11 '16 at 16:34

3 Answers3

4

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.

lab bhattacharjee
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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 $$ 4\cdot11\cdot12-13^2 $$

egreg
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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 $c=\sqrt{12}$.

Ángel Mario Gallegos
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