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 $\dfrac{{(aq + br + cp)(ar + bp + cq)}}{{{{(ap + bq + cr)}^2}}} = ?$
A.$1$
B.$- 1$
C.$abc$
D.$pqr$

Answer 1

First we have to define what the terms we need to solve the problem are.
First, complex numbers are the real and imaginary combined numbers as in the form of $z = x + iy$, where $x$ and $y$ are the real numbers and $i$ is the imaginary.
Imaginary $i$ can be also represented into the real values only if, ${i^2} = - 1$
Cube roots of unity $1,w,{w^2}$ , Cube roots of any number of the complex number
Then apply the formula, we can solve the requirement

Complete step by step answer:
It is given that,
Cube roots of unity
We know that $1,w,{w^2}$ are cube roots of unity
Thus, cube roots of any number ${k^3}$
So, will be $k,{\text{ }}kw,{\text{ }}k{w^2}$
Now, we need to find the value of the above condition
We know that,
$1,w,{w^2}$ are cube roots of unity
Now substituting these values for $p,q,r$ respectively
And using ${w^3} = {\text{ }}1$ and also ${w^4} = {\text{ w}}$ ${w^5} = {w^2}$
First, we apply the cube roots of unity in numerator
$(aq + br + cp)(ar + bp + cq)$
Therefore
Numerator becomes $= {k^2}(aw + b{w^{(2)}} + c)(a{w^2} + b + cw)$
${K^2}$ is the constant value of cube roots
$= {k^2}(aw + b{w^{(2)}} + c)(a{w^2} + b + cw)$
We see that equation like $\left( {a + b + c} \right)\left( {a + b + c} \right)$ this format
So, we apply the formula for the above equation
We know that the general formula
${\left( {a + b + c} \right)^2} = \left( {a + {\text{ }}b + c} \right)\left( {a + b + c} \right)$
So, we on simplifying,
${\left( {a + b + c} \right)^2} = \left( {{a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca} \right)$
Now we apply for the general formula in the above equation
We know that,
${\left( {a + b + c} \right)^2} = \left( {{a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca} \right)$
Now substituting for,
$= {k^2}\left( {{a^2} + abw + ac{w^2} + abw + {b^2}{w^2} + bc{\text{ }}ac{w^2} + bc + {c^2}w} \right)$
Then we apply the cube roots of unity in the denominator
${(ap + bq + cr)^2}$
Therefore, the denominator becomes $= {k^2}(aw + b{w^{(2)}} + c)(a{w^2} + b + cw)$
${K^2}$ is the constant value of cube roots
We see that equation like $\left( {a + b + c} \right)\left( {a + b + c} \right)$ this format
So, we apply the formula for the above equation
We know that the general formula
${\left( {a + b + c} \right)^2} = \left( {a + b + c} \right)\left( {a + b + c} \right)$
So, we on simplifying,
${\left( {a + b + c} \right)^2} = \left( {{a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca} \right)$
Now we apply for the general formula in the above equation
We know that,
${\left( {a + b + c} \right)^2} = \left( {{a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca} \right)$
And $p = 1,q = w,r = {w^2}$
Now substituting for,
$= \left( {{k^2}\left( {{a^2} + {\text{ }}abw{\text{ }} + {\text{ }}ac{w^2} + {\text{ }}abw{\text{ }} + {\text{ }}{b^2}{w^2} + {\text{ }}bc{\text{ }}ac{w^2} + {\text{ }}bc{\text{ }} + {\text{ }}{c^2}w} \right)} \right)$
Then, substituting the numerator and denominator
Therefore,
$= \dfrac{{(aq + br + cp)(ar + bp + cq)}}{{{{(ap + bq + cr)}^2}}}$
$= \dfrac{{{k^2}({a^2} + abw + ac{w^2} + abw + {b^2}{w^2} + bcac{w^2} + bc + {c^2}w)}}{{{k^2}({a^2} + abw + ac{w^2} + abw + {b^2}{w^2} + bcac{w^2} + bc + {c^2}w)}}$
$= 1$
Thus observing, the numerator and denominator are the same.

Hence, the correct option is (A).

Note:
The sum of the cube of unity result is the cube root of unity is referred to as the cube root of one. It is defined as the number that can be raised to the power of $3$ and the result is one. The sum of the three cube roots of unity results is zero. Then the product of the two imaginary cube roots is one or, the product of three cube roots of unity is one.