Question

Find the value of ${{\left( 2+\omega +{{\omega }^{2}} \right)}^{3}}-{{\left( 1-3\omega +{{\omega }^{2}} \right)}^{3}}-{{\left( 1-3\omega +{{\omega }^{2}} \right)}^{3}}$?
(a) 125
(b) 126
(c) 128
(d) 129

Answer 1

We start solving the problem by assuming the given sum equal to S. We then make the necessary arrangements inside the cubes of the given sum and then make use of the result that 1, $\omega$, ${{\omega }^{2}}$ are the complex cubic roots of unity and $1+\omega +{{\omega }^{2}}=0$. We then make necessary calculations and make use of the result ${{\omega }^{3}}=1$. We then make the necessary calculations to get the required answer.

Complete step by step answer:
According to the problem, we are asked to find the value of ${{\left( 2+\omega +{{\omega }^{2}} \right)}^{3}}-{{\left( 1-3\omega +{{\omega }^{2}} \right)}^{3}}-{{\left( 1-3\omega +{{\omega }^{2}} \right)}^{3}}$.
Let us assume $S={{\left( 2+\omega +{{\omega }^{2}} \right)}^{3}}-{{\left( 1-3\omega +{{\omega }^{2}} \right)}^{3}}-{{\left( 1-3\omega +{{\omega }^{2}} \right)}^{3}}$.
$\Rightarrow S={{\left( 1+1+\omega +{{\omega }^{2}} \right)}^{3}}-{{\left( -4\omega +1+\omega +{{\omega }^{2}} \right)}^{3}}-{{\left( -4\omega +1+\omega +{{\omega }^{2}} \right)}^{3}}$ ---(1).
We know that 1, $\omega$, ${{\omega }^{2}}$ are the complex cubic roots of unity. So, we have $1+\omega +{{\omega }^{2}}=0$. Let us substitute this result in equation (1).
$\Rightarrow S={{\left( 1+0 \right)}^{3}}-{{\left( -4\omega +0 \right)}^{3}}-{{\left( -4\omega +0 \right)}^{3}}$.
$\Rightarrow S={{\left( 1 \right)}^{3}}-{{\left( -4\omega \right)}^{3}}-{{\left( -4\omega \right)}^{3}}$.
$\Rightarrow S=1-\left( -64{{\omega }^{3}} \right)-\left( -64{{\omega }^{3}} \right)$ ---(2).
We know that ${{\omega }^{3}}=1$. Let us substitute this result in equation (2).
$\Rightarrow S=1-\left( -64\left( 1 \right) \right)-\left( -64\left( 1 \right) \right)$.
$\Rightarrow S=1-\left( -64 \right)-\left( -64 \right)$.
$\Rightarrow S=1+64+64$.
$\Rightarrow S=129$.
So, we have found the value of the given sum ${{\left( 2+\omega +{{\omega }^{2}} \right)}^{3}}-{{\left( 1-3\omega +{{\omega }^{2}} \right)}^{3}}-{{\left( 1-3\omega +{{\omega }^{2}} \right)}^{3}}$ as 129.

So, the correct answer is “Option d”.

Note: Whenever we get this type of problems, we should try to make use of the fact that 1, $\omega$, ${{\omega }^{2}}$ are the complex cubic roots of unity such that $1+\omega +{{\omega }^{2}}=0$, ${{\omega }^{3}}=1$ which leads us to the required answer. We should not make calculation mistakes while solving this type of problem. We can solve this problem by expanding the given cubes and then making the addition and subtraction operations manually which requires precision in calculation. Similarly, we can also expect problems to find the value of ${{\left( 2+\omega +{{\omega }^{2}} \right)}^{3}}-{{\left( 1-3\omega +{{\omega }^{2}} \right)}^{3}}-{{\left( 1+\omega -3{{\omega }^{2}} \right)}^{3}}$.