Question

If \[\omega \left( \ne 1 \right)\] is a cube root of unity and \[{{\left( 1+\omega \right)}^{7}}=A+B\omega \], then A & B are respectively the numbers
(a). 0, 1
(b). 1, 1
(c). 1, 0
(d). -1, 1

Answer 1

Hint: As \[\omega \] is cube root of unity so use the property of it to find the value \[{{\left( 1+\omega \right)}^{7}}\] by using fact that \[1+\omega +{{\omega }^{2}}=0\] or \[\left( 1+\omega \right)=-{{\omega }^{2}}\]. So, \[{{\left( 1+\omega \right)}^{7}}\] is \[{{\left( -{{\omega }^{2}} \right)}^{7}}\] or \[-{{\omega }^{14}}\]. After that multiply it by \[{{\omega }^{15}}\] or 1 as the value will not be changed because \[{{\omega }^{3}}\] is 1. Hence compare to get find values of A and B.

Complete step-by-step answer:

In the question we are given that wise cube root of unity and also if \[{{\left( 1+\omega \right)}^{7}}\] value is represented by \[A+B\omega \] then we have to find the value of A & B respectively.
In order to find the cube roots of unity we need to factorize the following cubic equation:
\[\Rightarrow \]\[{{x}^{3}}-1=0\]
Now we know that, \[{{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)\].
So, we can re – write the above equation as,
\[\Rightarrow \left( x-1 \right)\left( {{x}^{2}}+x+1 \right)=0\]
The given above equation gives one cube root or value of x as 1 while we have to find other two from the equation,
\[\Rightarrow {{x}^{2}}+x+1\]
Now we can solve it using formula,
\[\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
If the given quadratic equation is \[a{{x}^{2}}+bx+c=0\].
Here the given is \[{{x}^{2}}+x+1\]. So, the value of a, b, c is 1.
So, \[x=\dfrac{-1\pm \sqrt{{{\left( 1 \right)}^{2}}-4\times 1\times 2}}{2\times 1}\]
Or, \[x=\dfrac{-1\pm \sqrt{-3}}{2}\]
Now as we know that value of \[\sqrt{-1}\] is i.
So, the value of x is \[\dfrac{-1\pm \sqrt{3i}}{2}\].
If one of the found values of x other than 1this considered as \[\omega \] then other will be \[{{\omega }^{2}}\].
As \[\omega \] is the root of the quadratic equation \[{{x}^{2}}+x+1\]. So, if we substitute x as \[\omega \] we get,
\[\Rightarrow {{\omega }^{2}}+\omega +1=0\]
Now in the given question it was given that \[{{\left( 1+\omega \right)}^{7}}\]. So, as we know that \[{{\omega }^{2}}+\omega +1=0\].
\[\Rightarrow {{\omega }^{2}}+\omega +1=0\]
\[\Rightarrow \omega +1=-{{\omega }^{2}}\]
So, the value of \[\omega +1\] can be written as \[-{{\omega }^{2}}\].
Hence \[{{\left( 1+\omega \right)}^{7}}\] can be written as \[{{\left( -{{\omega }^{2}} \right)}^{7}}\] or \[-{{\omega }^{2\times 7}}\] or \[-{{\omega }^{14}}\].
Now as we know that \[{{\omega }^{3}}=1\] so if we multiply \[-{{\omega }^{14}}\] by \[{{\omega }^{3}}\] as 5 is a multiple of 3, therefore the answer will not change.
So, \[{{\omega }^{-14}}\times {{\omega }^{15}}\] is equal to \[\omega \].
So, the value of \[{{\left( 1+\omega \right)}^{7}}\] is \[\omega \].
Now we were given that, \[{{\left( 1+\omega \right)}^{7}}=A+B\omega \] and the value of \[{{\left( 1+\omega \right)}^{7}}\] is \[\omega \]. So,
\[\omega =A+B\omega \].
Now on comparing we can say that the value of A is 0 and B is 1.
Hence the correct option is (a).

Note: Here \[\omega \] is a complex number which is considered as the cube root of unity which is generally used to find the values of higher powers of \[\omega \]. The value of \[\omega \] is \[\dfrac{-1\pm \sqrt{3i}}{2}\] as it is said that the value of \[\dfrac{-1\pm \sqrt{3i}}{2}\] is 1.