Question

If $\omega$ is the complex cube root of unity then, show that $\left( {a - b} \right)\left( {a - b\omega } \right)\left( {a - b{\omega ^2}} \right) = {a^3} - {b^3}$.

Answer 1

Here in this question for showing the equality of the equations we will use the left hand side of the equation. So by using the LHS and solving the braces by the multiplication we can get to the values of the RHS easily.

Complete step by step answer:
Here, we have an equation given as $\left( {a - b} \right)\left( {a - b\omega } \right)\left( {a - b{\omega ^2}} \right) = {a^3} - {b^3}$
So, by taking the LHS of the equation we will get the equation as
$\Rightarrow \left( {a - b} \right)\left( {a - b\omega } \right)\left( {a - b{\omega ^2}} \right)$
Now on multiplying, we will get the equation as
$\Rightarrow \left( {{a^2} - ab\omega - ab - {b^2}\omega } \right)\left( {a - b{\omega ^2}} \right)$
And on multiplying furthermore, we will get the equation as
$\Rightarrow \left( {{a^3} - {a^2}b\omega - {a^2}b\iota {\omega ^2} + a{b^2}{\omega ^3} - {a^2}b + a{b^2}{\omega ^2} + a{b^2}\omega - {b^3}{\omega ^3}} \right)$
Since, we know that ${\omega ^3} = 1\,\,\& \,\,\left[ {\omega + {\omega ^2} = - 1} \right]$
Now on solving it furthermore, we get
$\Rightarrow {a^3} - {a^2}b\left( {\omega + {\omega ^2}} \right) + a{b^2}\left( {\omega + {\omega ^2}} \right) + a{b^2} - {b^3} - {a^2}b$
And on expanding the equation, we will get the equation as
$\Rightarrow {a^3} - {a^2}b + a{b^2} + a{b^2} - {b^3} - {a^2}b$
Since, the same term will cancel each other. Therefore we will get the values as ${a^3} - {b^3}$.
Hence, it is proved that the LHS is equal to the RHS.

Note:
For solving this type of question, either we can go with the LHS and RHS and by solving it, we will get the answer. But also by taking the RHS we can solve this route easily. So by using the formula which is given by ${\omega ^3} = 1\,\,\& \,\,\left[ {\omega + {\omega ^2} = - 1} \right]$. We can easily solve such questions.