Question

Let $\omega$ being the cube root of unity then prove that ${{\left( 3+3\omega +5{{\omega }^{2}} \right)}^{6}}-{{\left( 2+6\omega +2{{\omega }^{2}} \right)}^{3}}=0$ .

Answer 1

Hint: Use the property that ${{\omega }^{3n}}=1$ and $1+\omega +{{\omega }^{2}}=0$ , where $\omega$ represents cube root of unity. Start with the left-hand side of the equation given in the figure but try to avoid the binomial expansion as much as possible to make the solution easier and less complex.

Complete step-by-step answer:
Before starting with the solution, let us discuss some of the formulas related to the cube root of unity.
The important formulas include:
${{\omega }^{3n}}=1$
$1+\omega +{{\omega }^{2}}=0$
Now let us start with the solving of the left-hand side of the equation given in the question.
${{\left( 3+3\omega +5{{\omega }^{2}} \right)}^{6}}-{{\left( 2+6\omega +2{{\omega }^{2}} \right)}^{3}}$
$={{\left( 3+3\omega +3{{\omega }^{2}}+2{{\omega }^{2}} \right)}^{6}}-{{\left( 2+2\omega +2{{\omega }^{2}}+4\omega \right)}^{3}}$
Now if we take 3 common from the first three terms of the first bracket and 2 common from the first three terms of the second bracket, we get
${{\left( 3\left( 1+\omega +{{\omega }^{2}} \right)+2{{\omega }^{2}} \right)}^{6}}-{{\left( 2\left( 1+\omega +{{\omega }^{2}} \right)+4\omega \right)}^{3}}$
When we use the property $1+\omega +{{\omega }^{2}}=0$ , we get
${{\left( 3\times 0+2{{\omega }^{2}} \right)}^{6}}-{{\left( 2\times 0+4\omega \right)}^{3}}$
$={{\left( 0+2{{\omega }^{2}} \right)}^{6}}-{{\left( 0+4\omega \right)}^{3}}$
$={{2}^{6}}{{\omega }^{6\times 2}}-{{4}^{3}}{{\omega }^{3}}$
Now we will use the identity ${{\omega }^{3n}}=1$ . On doing so, we get
${{2}^{6}}{{\omega }^{3\times 4}}-{{4}^{3}}{{\omega }^{3}}$
$={{2}^{6}}\times 1-{{\left( {{2}^{2}} \right)}^{3}}\times 1$
$={{2}^{6}}-{{2}^{6}}=0$
So, we have shown that left-hand side of the equation ${{\left( 3+3\omega +5{{\omega }^{2}} \right)}^{6}}-{{\left( 2+6\omega +2{{\omega }^{2}} \right)}^{3}}=0$ is equal to right-hand side. Hence, we can say that we have proved that ${{\left( 3+3\omega +5{{\omega }^{2}} \right)}^{6}}-{{\left( 2+6\omega +2{{\omega }^{2}} \right)}^{3}}=0$

Note: Don’t get confused and apply the properties of the cube root of unity to all the places wherever you see the symbol $\omega$ , as the symbol has different meanings in different chapters and topics. So, be sure that you use the above mentioned properties with $\omega$ only if it is mentioned that $\omega$ represents the cube root of unity. Also, be careful while opening the brackets and multiplying the signs. Often, students commit mistakes while opening the brackets.