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.