Question

If $1,\omega ,{{\omega }^{2}}$are cube root of unity, then find the value of $\left( 1+3\omega +{{\omega }^{2}} \right)+{{\left( 1+3\omega -{{\omega }^{2}} \right)}^{4}}$ .

Answer 1

Hint: The given problem is related to cube roots of unity. We will solve this question using the following properties of cube roots of unity:
(i). $1+\omega +{{\omega }^{2}}=0$
(ii). ${{\omega }^{3}}=1$

Complete step-by-step answer:

To proceed with the solution, firstly, we will simplify the given expression. The given expression is $\left( 1+3\omega +{{\omega }^{2}} \right)+{{\left( 1+3\omega -{{\omega }^{2}} \right)}^{4}}$. We can split $3\omega$ as $\omega +2\omega$ and $-{{\omega }^{2}}$ in the second term as $-2{{\omega }^{2}}+{{\omega }^{2}}$. So, the expression changes to $\left( 1+\omega +{{\omega }^{2}}+2\omega \right)+{{\left( 1+\omega +{{\omega }^{2}}+2\omega -2{{\omega }^{2}} \right)}^{4}}$ . Now, we know $1+\omega +{{\omega }^{2}}=0$ .
$\Rightarrow \left( 1+\omega +{{\omega }^{2}}+2\omega \right)+{{\left( 1+\omega +{{\omega }^{2}}+2\omega -2{{\omega }^{2}} \right)}^{4}}=\left( 0+2\omega \right)+{{\left( 0+2\omega -2{{\omega }^{2}} \right)}^{4}}$.
$\Rightarrow \left( 1+3\omega +{{\omega }^{2}} \right)+{{\left( 1+3\omega -{{\omega }^{2}} \right)}^{4}}=2\omega +{{\left( 2\omega -2{{\omega }^{2}} \right)}^{4}}$
Now, we will take 2$\omega$ common from the second term. So, on taking 2$\omega$ common, we get:
$\left( 1+3\omega +{{\omega }^{2}} \right)+{{\left( 1+3\omega -{{\omega }^{2}} \right)}^{4}}=2\omega +{{\left( 2\omega \left( 1-\omega \right) \right)}^{4}}$
$\Rightarrow \left( 1+3\omega +{{\omega }^{2}} \right)+{{\left( 1+3\omega -{{\omega }^{2}} \right)}^{4}}=\left( 2\omega \right)+16{{\omega }^{4}}{{\left( 1-\omega \right)}^{4}}.......(i)$
Now, we know that the expansion of ${{\left( 1-x \right)}^{4}}={{x}^{4}}-4{{x}^{3}}+6{{x}^{2}}-4x+1$. So, the expansion of ${{\left( 1-\omega \right)}^{^{4}}}$will be ${{\left( 1-\omega \right)}^{^{4}}}={{\omega }^{4}}-4{{\omega }^{3}}+6{{\omega }^{2}}-4\omega +1$
Now, we know that the value of ${{\omega }^{3}}=1$ .
$\Rightarrow {{\left( 1-\omega \right)}^{^{4}}}=\omega -4+6{{\omega }^{2}}-4\omega +1$
$\Rightarrow {{\left( 1-\omega \right)}^{^{4}}}=-3\omega -3+6{{\omega }^{2}}$
Taking -3 common, we get:
${{\left( 1-\omega \right)}^{^{4}}}=-3\left( 1+\omega -2{{\omega }^{2}} \right)$
$\Rightarrow {{\left( 1-\omega \right)}^{^{4}}}=-3\left( 1+\omega +{{\omega }^{2}}-3{{\omega }^{2}} \right)$
Now, we know, $1+\omega +{{\omega }^{2}}=0$
$\Rightarrow {{\left( 1-\omega \right)}^{4}}=-3\left( 0-3{{\omega }^{2}} \right)$
$\Rightarrow {{\left( 1-\omega \right)}^{4}}=9{{\omega }^{2}}$
On substituting ${{\left( 1-\omega \right)}^{4}}=9{{\omega }^{2}}$ in equation (i), we get:
$\Rightarrow \left( 1+3\omega +{{\omega }^{2}} \right)+{{\left( 1+3\omega -{{\omega }^{2}} \right)}^{4}}=\left( 2\omega \right)+16{{\omega }^{4}}\times 9{{\omega }^{2}}$
=$\Rightarrow \left( 1+3\omega +{{\omega }^{2}} \right)+{{\left( 1+3\omega -{{\omega }^{2}} \right)}^{4}}=\left( 2\omega \right)+144{{\omega }^{6}}$
$\Rightarrow \left( 1+3\omega +{{\omega }^{2}} \right)+{{\left( 1+3\omega -{{\omega }^{2}} \right)}^{4}}=\left( 2\omega \right)+144{{\left( {{\omega }^{3}} \right)}^{2}}$
$\Rightarrow \left( 1+3\omega +{{\omega }^{2}} \right)+{{\left( 1+3\omega -{{\omega }^{2}} \right)}^{4}}=\left( 2\omega \right)+144$
Hence, the value of $\left( 1+3\omega +{{\omega }^{2}} \right)+{{\left( 1+3\omega -{{\omega }^{2}} \right)}^{4}}$comes out to be $144+2\omega$

Note: While making substitution, take care of the sign. Sign mistakes are very common and students can get a wrong answer even due to a single sign mistake. So, students should perform calculations and substitutions very carefully.