Question

The value of $(a+b\omega +c{\omega }^2{)}^3+(a+b{\omega }^2+c\omega {)}^3$ is/are
[Here $\omega$ is a complex cube root of unity]

• A. $(2a-b-c)(2b-a-c)(2c-a-b)$
• B. $9abc,\text{if}a+b+c=0$
• C. $27abc,\text{if}a+b+c=0$
• D. $(2a+b+c)(2b-a-c)(2c-a-b)$

Answer 1

Answer: (A), (C)

The correct options are
A $(2a-b-c)(2b-a-c)(2c-a-b)$
C $27abc,\text{if}a+b+c=0$

Let $a+b\omega +c{\omega }^2=x\text{and}a+b{\omega }^2+c\omega =y$

$\therefore (a+b\omega +c{\omega }^2{)}^3+(a+b{\omega }^2+c\omega {)}^3=x^3+y^3=(x+y)(x+\omega y)(x+{\omega }^{2}y)$

Now,
$x+y=(a+b\omega +c{\omega }^2)+(a+b{\omega }^2+c\omega )$
$=2a+b(\omega +{\omega }^2)+c(\omega +{\omega }^2)$
$=(2a-b-c)\cdots \left(1\right)$

$x+\omega y=(a+b\omega +c{\omega }^2)+\omega (a+b{\omega }^2+c\omega )$
$=(1+\omega )a+(1+\omega )b+2{\omega }^{2}c$
$={\omega }^2(2c-a-b)\cdots \left(2\right)$

$x+{\omega }^{2}y=(a+b\omega +c{\omega }^2)+{\omega }^2(a+b{\omega }^2+c\omega )$
$=\omega (2b-a-c)\cdots \left(3\right)$

Hence from $\left(1\right),\left(2\right)$ and $\left(3\right)$
$x^3+y^3=(2a-b-c)(2c-a-b)(2b-a-c)\cdots \left(4\right)$

If
$a+b+c=0\Rightarrow b+c=-a,c+a=-b,a+b=-c$
Putting these values in the equation $\left(4\right),$ we get
$x^3+y^3=\left(3a\right)\left(3b\right)\left(3c\right)=27abc$