Question

If $\omega $ is a cube root of unity, then the value of ${{\left( 1+\omega +{{\omega }^{2}} \right)}^{5}}+{{\left( 1+\omega -{{\omega }^{2}} \right)}^{5}}$ is
[a] $16\omega $
[b] $32\omega $
[c] $48\omega $
[d] $-32\omega $

Answer 1

Hint: Use the fact that $\omega $ is a complex cube root of unity if and only if $\omega $ satisfies the quadratic equation $1+x+{{x}^{2}}=0$. Also since $\omega $ is a cube root of unity, we have ${{\omega }^{3}}=1$. Simplify and solve the expression using these two properties.

Complete step-by-step solution -

We have $\omega $ is a cube root of unity, then $\omega $ is a root of the cubic ${{x}^{3}}=1$
Hence we have ${{x}^{3}}-1=0$
Using ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$, hence we have
${{x}^{3}}-1=\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)$
Since $\omega $ is not purely real, we have $\omega \ne 1$
Hence $\omega $ satisfies ${{x}^{2}}+x+1=0$
Hence we have $1+\omega +{{\omega }^{2}}=0$.
Hence $1+\omega =-{{\omega }^{2}}$
Hence we have ${{\left( 1+\omega -{{\omega }^{2}} \right)}^{5}}={{\left( -2{{\omega }^{2}} \right)}^{5}}=-32{{\omega }^{10}}=-32{{\left( {{\omega }^{3}} \right)}^{3}}\omega =-32\omega $
Hence we have ${{\left( 1+\omega +{{\omega }^{2}} \right)}^{5}}+{{\left( 1+\omega -{{\omega }^{2}} \right)}^{5}}=0-32\omega =-32\omega $
Hence option [d] is correct.

Note: If $\alpha $ is an nth root of unity, then we have $1+\alpha +{{\alpha }^{2}}+\cdots +{{\alpha }^{n-1}}=0$, if $\alpha $ is not purely real.
Proof: Since $\alpha $ is an nth root of unity, we have ${{\alpha }^{n}}=1$.
Now since $\alpha $ is not purely real $\alpha -1\ne 0$.
Observe that $1,\alpha .{{\alpha }^{2}},\cdots ,{{\alpha }^{n-1}}$ for a geometric progression with a = 1 and $r=\alpha $.
We know that in a geometric progression ${{S}_{n}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$
Using the above result, we have
${{S}_{n}}=\dfrac{1\left( {{\alpha }^{n}}-1 \right)}{\alpha -1}$
Since ${{\alpha }^{n}}=1$, we have
${{S}_{n}}=\dfrac{\left( 1-1 \right)}{\alpha -1}=0$
Hence, we have $1+\alpha +{{\alpha }^{2}}+\cdots +{{\alpha }^{n-1}}=0$
Hence proved.
[2] Put n = 3 in the above formula and $\alpha =\omega $, we get
$1+\omega +{{\omega }^{2}}=0$, which is the same as obtained above.