Question

If $x=a+b+c,y=a\alpha +b\beta +c$ and $z=a\beta +b\alpha +c$,where $\alpha$ and $\beta$ are imaginary cube roots of unity,then $xyz$

• A. $2(a^3+b^3+c^3)$
• B. $2(a^3-b^3-c^3)$
• C. $(a^3+b^3+c^3)-3abc$
• D. $a^3-b^3-c^3$

Answer 1

Answer: (C)

$(a^3+b^3+c^3)-3abc$

$\because x=(a+b+c)y=(a\alpha +b\beta +c)z=(a\beta +b\alpha +c)$

also $\alpha$ and $\beta$ are imaginary cube roots of unity

Let $\alpha =\omega and\beta ={\omega }^2$

now,

$xyz=(a+b+c)(a\alpha +b\beta +c)(a\beta +b\alpha +c)=\alpha \beta (a^3+a{b}^2+a^{2}b+b^3+a^{2}c+b^{2}c)+({\alpha }^2+{\beta }^2)(a^{2}b+a{b}^2+abc)+(\alpha +\beta )(a^{2}c+2abc+b^{2}c+a{c}^2+b{c}^2)+(a{c}^2+b{c}^2+c^3)But,\omega {.\omega }^2={\omega }^3=1and,{\omega }^2+{\omega }^4={\omega }^2+\omega .{\omega }^3={\omega }^2+\omega .1={\omega }^2+\omega \because {\omega }^2+\omega +1=0\Rightarrow {\omega }^2+\omega =-1\therefore xyz=\omega {.\omega }^2(a^3+a{b}^2+a^{2}b+b^3+a^{2}c+b^{2}c)+({\omega }^2+{\omega }^4)(a^{2}b+a{b}^2+abc)+({\omega }^2+\omega )(a^{2}c+2abc+b^{2}c+a{c}^2+b{c}^2)+(a{c}^2+b{c}^2+c^3)Byputtingtheabovevaluewegetxyz=(a^3+b^3+c^3)-3abc$