Question

$1,\omega ,{\omega }^2$ are cube roots of unity, then the value of $(a+b+c)(a+b\omega +c{\omega }^2)(a+b{\omega }^2+c\omega )=$

• A. $a^3+b^3+c^3$
• B. $a^3+b^3+c^3+3abc$
• C. $a^3+b^3+c^3-3abc$
• D. $a^3+b^3+c^3-abc$

Answer 1

The correct option is C $a^3+b^3+c^3-3abc$

$(a+b+c)\cdot (a+b\omega +c{\omega }^2)\cdot (a+b{\omega }^2+c\omega )$

Expanding the brackets,

$=(a^2+ab\omega +ac{\omega }^2+ba+b^2\omega +bc{\omega }^2+ca+bc\omega +c{\omega }^2)\cdot (a+b{\omega }^2+c\omega )$

$=(a^2+ab+ca+(ab+b^2+bc)\omega +(ac+bc+c^2){\omega }^2)\cdot (a+b{\omega }^2+c\omega )$

$=a^3+a^{2}b+c{a}^2+(a^{2}b+b^{2}a+abc)\omega +(a^{2}c+abc+a{c}^2){\omega }^2+(a^{2}b+a{b}^2+ca){\omega }^{20}+(a{b}^2+b^3+b^{2}c){\omega }^2+(abc+b^{2}c+b{c}^2){\omega }^4+(a^{2}c+abc+a{c}^2)\omega +(abc+b^{2}c+b{c}^2){\omega }^2+(a{c}^2+b{c}^2+c^3){\omega }^3$

$=a^3+a^{2}b+c{a}^2+a{c}^2+b{c}^2+c^2+a{b}^2+b^3+b^{2}c+(a^{2}c+abc+a{c}^2+abc+b{c}^2+b{c}^2+a^{2}b+b{a}^2+abc)+(abc+b^{2}c+b{c}^2+a^{2}b+a{b}^2+cab+a^{2}c+abc+a{c}^2)$

$=a^3+b^3+c^3+(a^{2}c+a{c}^2+b{c}^2+b^{2}c+a^{2}b+a{b}^2)(1+\omega +{\omega }^2)+\left(3abc\right)(\omega +{\omega }^2)$

$=a^3+b^3+c^3+0+\left(3abc\right)(-1)(\because 1+\omega +{\omega }^2=0)$

$=a^3+b^3+c^3-3abc$

Hence, option C.