Question

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

• A. a² + b²
• B. a³ + b³
• C. a³ b³
• D. a³ - b³

Answer 1

The correct option is B a³ + b³

CONVENTIONAL APPROACH:
If $x=a+by=a\alpha +b\beta$ and $z=\alpha \beta +\beta \alpha$
Then $xyz=(a+b)(aw+b{w}^2)(a{w}^2+bw)$,where $\alpha =w$ and $\beta =w^2=(a+b)(a^2+ab{w}^2+abw+b^2)$
$=(a+b)(a^2-ab+b^2)=a^3+b^3$

This is a variable to variable question i.e. the question is invariables and the options
are in variables.So, we can assume and substitute any value for the variables.

Tricks: Put $a=b=2$
Then $x=4,y=2(w+w^2)=-2$ and $z=2(w^2+w)=-2$
$\because xyz=4(-2)(-2)=16$ and $\left(b\right)$ i.e. $a^3+b^3=16.$

This method works on elimination. $\left(B\right)$ is the correct option because $\left(A\right),\left(B\right)$ and $\left(D\right)$ are
not equal to 16.