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 $2x+1=i\sqrt{3}$
\[\begin{align}
  & {{\left( 2x+1 \right)}^{2}}={{\left( i\sqrt{3} \right)}^{2}} \\
 & \Rightarrow 4{{x}^{2}}+4x+1=3{{i}^{2}} \\
\end{align}\]
We know that for imaginary value $i$, we get ${{i}^{2}}=-1$
\[\begin{align}
  & 4{{x}^{2}}+4x+1=-3 \\
 & \Rightarrow 4{{x}^{2}}+4x+4=0 \\
 & \Rightarrow {{x}^{2}}+x+1=0 \\
\end{align}\]
We now multiply $x-1$ to both sides of the equation and get
\[\begin{align}
  & \left( x-1 \right)\left( {{x}^{2}}+x+1 \right)=0 \\
 & \Rightarrow {{x}^{3}}-1=0 \\
\end{align}\]
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
\[\begin{align}
  & {{\left( 1-{{x}^{2}}+x \right)}^{6}}-{{\left( 1+{{x}^{2}}-x \right)}^{6}} \\
 & ={{\left( 1-{{\omega }^{2}}+\omega \right)}^{6}}-{{\left( 1+{{\omega }^{2}}-\omega \right)}^{6}} \\
\end{align}\]
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.