Question

If $2x=-1+i\sqrt{3}$, then the value of ${{\left( 1-{{x}^{2}}+x \right)}^{6}}-{{\left( 1+{{x}^{2}}-x \right)}^{6}}$ is
A. 32
B. $-64$
C. 64
D. 0

Answer 1

Hint : We first find the speciality about the given equation $2x=-1+i\sqrt{3}$. We take squares and multiply with $x-1$. We use the value of $x=\omega$ as the imaginary cube root of unity. We put the value in the expression and find the value using the identities of $1+\omega +{{\omega }^{2}}=0;{{\omega }^{3}}=1$.

Complete step-by-step answer :
We first simplify the equation $2x=-1+i\sqrt{3}$ by squaring both sides. We get $2x+1=i\sqrt{3}$

& {{\left( 2x+1 \right)}^{2}}={{\left( i\sqrt{3} \right)}^{2}} \\\ & \Rightarrow 4{{x}^{2}}+4x+1=3{{i}^{2}} \\\ \end{aligned}$$ We know that for imaginary value $i$, we get ${{i}^{2}}=-1$ $$\begin{aligned} & 4{{x}^{2}}+4x+1=-3 \\\ & \Rightarrow 4{{x}^{2}}+4x+4=0 \\\ & \Rightarrow {{x}^{2}}+x+1=0 \\\ \end{aligned}$$ We now multiply $x-1$ to both sides of the equation and get $$\begin{aligned} & \left( x-1 \right)\left( {{x}^{2}}+x+1 \right)=0 \\\ & \Rightarrow {{x}^{3}}-1=0 \\\ \end{aligned}$$ Therefore, we get the cube root of unity as the value of $x$. We know it has three roots $$1,\omega ,{{\omega }^{2}}$$. As we have $x=\dfrac{-1+i\sqrt{3}}{2}$, we can take it as any of the imaginary roots. We take $x=\omega $. With the value of $\omega $, we know the identities involved and they are $$1+\omega +{{\omega }^{2}}=0;{{\omega }^{3}}=1$$. We put the value of $x=\omega $ in $${{\left( 1-{{x}^{2}}+x \right)}^{6}}-{{\left( 1+{{x}^{2}}-x \right)}^{6}}$$ and get $$\begin{aligned} & {{\left( 1-{{x}^{2}}+x \right)}^{6}}-{{\left( 1+{{x}^{2}}-x \right)}^{6}} \\\ & ={{\left( 1-{{\omega }^{2}}+\omega \right)}^{6}}-{{\left( 1+{{\omega }^{2}}-\omega \right)}^{6}} \\\ \end{aligned}$$ Using the identity $$1+\omega +{{\omega }^{2}}=0;{{\omega }^{3}}=1$$, we get $$\ ={\left( {1 - {\omega ^2} + \omega } \right)^6} - {\left( {1 + {\omega ^2} - \omega } \right)^6} \\\ {\left( { - \omega - {\omega ^2} - {\omega ^2} + \omega } \right)^6} - {\left( { - \omega - {\omega ^2} + {\omega ^2} - \omega } \right)^6} \\\ = {\left( { - {\omega ^2} - {\omega ^2}} \right)^6} - {\left( { - \omega - \omega } \right)^6} \\\ = 64{\omega ^{12}} - 64{\omega ^6} \\\ = 64{\left( {{\omega ^3}} \right)^4} - 64{\left( {{\omega ^3}} \right)^2} \\\ = 64 - 64 \\\ = 0 \;

Therefore, the value of ${{\left( 1-{{x}^{2}}+x \right)}^{6}}-{{\left( 1+{{x}^{2}}-x \right)}^{6}}$ is 0. The correct option is D.
So, the correct answer is “Option D”.

Note: We can also solve the problem assuming the value of $x={{\omega }^{2}}$. We can take the imaginary value of $x=\dfrac{-1+i\sqrt{3}}{2}$ as any of $\omega ,{{\omega }^{2}}$ as the square value gives the other imaginary root of $x=\dfrac{-1-i\sqrt{3}}{2}$. Also, we need to remember that as we multiplied the term $x-1$, it gives us the real root.