Question

If $z(z\ne 0)$ p,q,r are distinct cube roots of a complex number and $a$, b,c are complex numbers such that $ap+bq+cr\ne 0$

then $\frac{(aq+br+cp)(ar+bp+cq)}{(ap+bq+cr{)}^2}=?$

• A. $1$
• B. $-1$
• C. $abc$
• D. $pqr$

Answer 1

The correct option is A $1$

We know that $1,w$ and $w^2$ are cube roots of unity.
Thus, cube roots of any number $k^3$ will be $k,kw$ and $k{w}^2.$
Now substituting these values for $p,q$ and $r$ respectively and using $w^3=1,$
numerator becomes $k^2\times (aw+b{w}^2+c)(a{w}^2+b+cw)=k^2(a^2+abw+ac{w}^2+abw+b^2{w}^2+bc+ac{w}^2+bc+c^{2}w)$
Denominator becomes $k^2(c^{2}w+a^2+abw+ac{w}^2+abw+b^2{w}^2+bc+ac{w}^2+bc)$
Thus observing, the numerator and denominator are the same.
Hence answer becomes 1.