Question

Simplify the following cyclic expressions.
(i) ${\sum }_{a,b,c}(b-c)(b+c)$ (ii) ${\left({\sum }_{a,b,c}a\right)}^2-\left({\sum }_{a,b,c}{a}^2\right)$ (iii) ${\sum }_{a,b,c}(a+1{)}^3(b^2-c^2)$

Answer 1

(i) ${\sum }_{a,b,c}(b-c)(b+c)$

cyclic expressions refers to substitution of variables in cyclic manner.

Here, $a,b,c$ are variables & the two expressions are multiplied.

So, ${\sum }_{a,b,c}(b-c)(b+c)=(b-c)(b+c)+(c-a)(c+a)+(a-b)(a+b)$

$=b^2-c^2+c^2-a^2+a^2-b^2$

$=0$

(ii) ${\left({\sum }_{a,b,c}a\right)}^2-\left({\sum }_{a,b,c}{a}^2\right)$

$\Rightarrow {(a+b+c)}^2-(a^2+b^2+c^2)$

$\Rightarrow a^2+b^2+c^2+2ab+2ca+2bc-a^2-b^2-c^2$

$\Rightarrow 2(ab+bc+ca)$

(iii) ${\sum }_{a,b,c}{(a+1)}^3(b^2-c^2)$

$\Rightarrow {(a+1)}^3(b^2-c^2)+{(b+1)}^3(c^2-a^2)+{(c+1)}^3(a^2-b^2)$

$\Rightarrow (a^3+3a^2+3a+1)(b^2-c^2)+(b^3+3b^2+3b+1)(c^2-a^2)+(c^3+3c^2+3c+1)(a^2-b^2)$

$\Rightarrow a^3{b}^2-a^3{c}^2+3a^2{b}^2-3a^2{c}^2+3{ab}^2-3{ac}^2+b^2-c^2+b^3{c}^2-b^3{a}^2+3b^2{c}^2-3b^2{a}^2$

$+3{bc}^2-3{ba}^2+c^2-a^2+c^3{a}^2-c^3{b}^2+3c^2{a}^2-3c^2{b}^2+3{ca}^2-3{cb}^2+a^2-b^2$

$\Rightarrow a^3{b}^2-a^3{c}^2+3{ab}^2-3{ac}^2+b^3{c}^2-b^3{a}^2+3{bc}^2-3{ba}^2+c^3{a}^2-c^3{b}^2+3{ca}^2-3{cb}^2$

$\Rightarrow (a^3+3a)(b^2-c^2)+(b^3+3b)(c^2-a^2)+(c^3+3c)(a^2-b^2)$