Question

If the value of x, y and z are given as $x = a + b$, $y = a\alpha + b\beta$ and $z = a\beta + b\alpha$, where $\alpha ,\beta$ complex cube are roots of unity, show that $xyz = {a^3} + {b^3}$.

Answer 1

Hint: In this question $\alpha = \omega ,\beta = {\omega ^2}$, firstly substitute the value of x, y and z into $xyz$ and then substitute the value of$\alpha ,\beta$. Use the properties of complex cube roots of unity like$1 + \omega + {\omega ^2} = 0, {\omega ^3} = 1$, to get the answer.

Complete step-by-step answer:

$x = a + b$, $y = a\alpha + b\beta$ and $z = a\beta + b\alpha$ where $\alpha ,\beta$ are the complex cube root of unity.
So if $\alpha ,\beta$ are the complex cube root of unity.
So let $\alpha = \omega ,\beta = {\omega ^2}$ and it satisfy the equation
$1 + \omega + {\omega ^2} = 0,{\omega ^3} = 1$........................... (1)
Now we have to prove
$xyz = {a^3} + {b^3}$
Proof –
Consider L.H.S
$\Rightarrow xyz$
Now substitute the value of x, y and z we have,
$\Rightarrow \left( {a + b} \right)\left( {a\alpha + b\beta } \right)\left( {a\beta + b\alpha } \right)$
Now simplify the above equation we have,
$\Rightarrow \left( {a + b} \right)\left( {{a^2}\alpha \beta + ab{\alpha ^2} + ab{\beta ^2} + {b^2}\alpha \beta } \right)$
$\Rightarrow {a^3}\alpha \beta + {a^2}b{\alpha ^2} + {a^2}b{\beta ^2} + a{b^2}\alpha \beta + {a^2}b\alpha \beta + a{b^2}{\alpha ^2} + a{b^2}{\beta ^2} + {b^3}\alpha \beta$
Now substitute $\alpha = \omega ,\beta = {\omega ^2}$ in this equation we have,
$\Rightarrow {a^3}{\omega ^3} + {a^2}b{\omega ^2} + {a^2}b{\omega ^4} + a{b^2}{\omega ^3} + {a^2}b{\omega ^3} + a{b^2}{\omega ^2} + a{b^2}{\omega ^4} + {b^3}{\omega ^3}$
Now substitute ${\omega ^3} = 1$ we have,
$\Rightarrow {a^3} + {a^2}b{\omega ^2} + {a^2}b\omega + a{b^2} + {a^2}b + a{b^2}{\omega ^2} + a{b^2}\omega + {b^3}$
Now take ${a^2}b$ common from second, third and fifth term and take common $a{b^2}$from fourth, fifth and seventh term we have.
$\Rightarrow {a^3} + {a^2}b\left( {1 + \omega + {\omega ^2}} \right) + a{b^2}\left( {1 + \omega + {\omega ^2}} \right) + {b^3}$
Now from equation (1) we have,
$\Rightarrow {a^3} + {a^2}b\left( 0 \right) + a{b^2}\left( 0 \right) + {b^3} = {a^3} + {b^3}$
= R.H.S
Hence proved.

Note: The complex cube root of unity refers to cube root of 1 that is ${\left( 1 \right)^{\dfrac{1}{3}}} = \left( {1,\omega ,{\omega ^2}} \right)$. It is advised to remember the basic cube root of unity properties, some of them are being mentioned above. The basic concept of cube root arises by the roots of quadratic equation${x^2} + x + 1 = 0$, so the roots of this are $\omega = \dfrac{{ - 1 \pm i\sqrt 3 }}{2}$.