Question

Simplify the following cyclic expressions.

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

• A. $a^2(c^3-b^3)+b^2(a^3-c^3)+c^2(b^3-a^3)$
• B. $-\left(a^2\right(c^3-b^3)+b^2(a^3-c^3)+c^2(b^3-a^3\left)\right)$
• C. $a^2(c^3+b^3)+b^2(a^3+c^3)+c^2(b^3+a^3)$
• D. None

Answer 1

The correct option is D None

Cyclic expressions refers to substitution of variables in cyclic manner.

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

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

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

$=(a^3+1+3a^2+3a)(b^2-c^2)+(b^3+1+3b^2+3b)(c^2-a^2)+(c^3+1+3c^2+3c)(a^2-b^2)[\because (x+y{)}^3=x^3+y^3+3x^{2}y+3x{y}^2]$

$=a^3{b}^2-a^3{c}^2+b^2-c^2+3a^2{b}^2-3a^2{c}^2+3a{b}^2-3a{c}^2+b^3{c}^2-b^3{a}^2+c^2-a^2+3b^2{c}^2-3b^2{a}^2+3b{c}^2-3b{a}^2+c^3{a}^2-c^3{b}^2+a^2-b^2+3c^2{a}^2-3c2b^2+3c{a}^2-3c{b}^2$

$=a^2(b^3-c^3)+b^2(a^3-c^3)+c^2(b^3-a^3)+3\left[ab\right(b-a)+bc(b-c)+ac(c-a\left)\right]$

Hence, ${\sum }_{a,b,c}(a+1{)}^3(b^2-c^2)$$=a^2(b^3-c^3)+b^2(a^3-c^3)+c^2(b^3-a^3)+3\left[ab\right(b-a)+bc(b-c)+ac(c-a\left)\right]$

So, $\text{D}$ is the correct option.