For those who don't know part 2 (omega)

$X=(\frac{9-7\omega}{9\omega^2-7}-\frac{5-3\omega^2}{5\omega-3})^8$

$=(\omega\frac{9-7\omega}{9-7\omega}-\omega^2\frac{5-3\omega^2}{5-3\omega^2})^8$

$=(\omega-\omega^2)^8$

$=[(\omega-\omega^2)^2]^4$

$=[\omega^2+\omega-2]^4$

$=(-3)^4=\boxed{81}$

Note :the solution is the first two lines only the other paragraphs are only for more understanding.

Because $\omega$ is one of the cubic roots of $1$ $\therefore { \omega }^{ 3 }=1$

$\boxed { X={ (\frac { 9-7\omega }{ 9{ \omega }^{ 2 }-7 } -\frac { 5-3{ \omega }^{ 2 } }{ 5\omega -3 } ) }^{ 8 }={ (\frac { 9{ \omega }^{ 3 }-7\omega }{ 9{ \omega }^{ 2 }-7 } -\frac { 5{ \omega }^{ 3 }-3{ \omega }^{ 2 } }{ 5\omega -3 } ) }^{ 8 }={ (\frac { \omega (9{ \omega }^{ 2 }-7) }{ 9{ \omega }^{ 2 }-7 } -\frac { { \omega }^{ 2 }(5\omega -3) }{ 5\omega -3 } ) }^{ 8 }={ (\omega -{ \omega }^{ 2 }) }^{ 8 }={ (\pm \sqrt { 3 } i) }^{ 8 }=81{ i }^{ 8 }=\boxed { 81 } }$

• For those who don't understand why ${ (\omega -{ \omega }^{ 2 }) }=\pm \sqrt { 3 } i$ $\quad \quad \omega \quad =\quad \frac { -1\pm \sqrt { 3 } i }{ 2 } \quad then\quad { \omega }^{ 2 }={ (\frac { -1\pm \sqrt { 3 } i }{ 2 } ) }^{ 2 }=\frac { -1\mp \sqrt { 3 } i }{ 2 }$ , so ${ (\omega -{ \omega }^{ 2 }) }$ could be ${ (\omega -{ \omega }^{ 2 }) }=(\frac { -1+\sqrt { 3 } i }{ 2 } )-(\frac { -1-\sqrt { 3 } i }{ 2 } )=\sqrt { 3 } i$ or could be ${ (\omega -{ \omega }^{ 2 }) }=(\frac { -1-\sqrt { 3 } i }{ 2 } )-(\frac { -1+\sqrt { 3 } i }{ 2 } )=-\sqrt { 3 } i$

$\therefore { (\omega -{ \omega }^{ 2 }) }=\pm \sqrt { 3 } i$

• For those who don't understand why $\omega =\frac { -1\pm \sqrt { 3 } i }{ 2 }$ and ${ \omega }^{ 2 }=\frac { -1\mp \sqrt { 3 } i }{ 2 }$

( copied from Kalpok Ethbrw solution )

${ X }^{ 3 }=1$ or $(x-1)(x^2+x+1)=0$

Either $x=1$ or, $x^2+x+1=0$ Then $x = \frac { - 1 \pm \sqrt{ -1^2 - 4.1.1 } } { 2.1}$ or, $x = \frac { - 1 \pm \sqrt{ -3 } } { 2}$ or, one root of x is $\frac{-1+\sqrt{3}.i} {2}.$ This is called $\omega$ if we take ${\omega }^{ 2 }$ We will discover it is another root of the equation. The root is $\frac{-1-\sqrt{3}.i} {2}.$

Note : we can also take $\omega =\frac { -1-\sqrt { 3 } i }{ 2 }$ then ${ \omega }^{ 2 }=\frac { -1+\sqrt { 3 } i }{ 2 }$

$\therefore\omega =\frac { -1\pm \sqrt { 3 } i }{ 2 }$ and ${ \omega }^{ 2 }=\frac { -1\mp \sqrt { 3 } i }{ 2 }$

Since $\omega$ and $\omega^2$ are complex cubic roots of $1$ :

$\quad \Rightarrow \omega^3 = 1$ and $\omega^2 + \omega + 1 = 0$

Therefore,

$\quad X = \left( \dfrac {9-7\omega}{9\omega^2-7} - \dfrac {5-3\omega^2}{5\omega - 3} \right)^8 = \left( \dfrac {9\omega^2-7\omega^3}{\omega^2(9\omega^2-7)} - \dfrac {5\omega-3\omega^3}{\omega(5\omega - 3)} \right)^8$

$\quad \quad = \left( \dfrac {9\omega^2-7}{\omega^2(9\omega^2-7)} - \dfrac {5\omega-3}{\omega(5\omega - 3)} \right)^8 = \left( \dfrac {1}{\omega^2} - \dfrac {1}{\omega} \right)^8$

$\quad \quad = \left( \dfrac {\omega^3}{\omega^2} - \dfrac {\omega^3}{\omega} \right)^8 = \left( \omega - \omega^2 \right)^8 = \left( (\omega - \omega^2)^2 \right)^4$

$\quad \quad = \left( \omega^2 - 2\omega^3+\omega^4 \right)^4 = \left( \omega^2 - 2 + \omega \right)^4 = \left( \omega^2 + \omega + 1 - 3 \right)^4$

$\quad \quad = \left( 0 - 3 \right)^4 = \boxed {81}$

$\quad \quad \quad \quad \quad \quad \quad \quad { (\frac { 9-7\omega }{ 9{ \omega }^{ 2 }-7 } -\frac { 5-3{ \omega }^{ 2 } }{ 5\omega -3 } ) }^{ 8 }\\ Multiplying\quad and\quad dividing\quad \omega \quad in\quad the\quad numerator\\ and\quad denominator\quad of\quad both\quad the\quad terms.\\ \quad =\quad { ((\omega )\frac { 9-7\omega }{ (\omega )9{ \omega }^{ 2 }-7(\omega ) } -\frac { (\omega )5-3{ \omega }^{ 2 }(\omega ) }{ 5\omega -3 } \times \frac { 1 }{ \omega } ) }^{ 8 }\\ \quad \\ =\quad { (\frac { 9-7\omega }{ 9{ \omega }^{ 3 }-7\omega } (\omega )-\frac { 5\omega -3{ \omega }^{ 3 } }{ 5\omega -3 } \times \frac { 1 }{ \omega } ) }^{ 8 }\\ But\quad { \omega }^{ 3 }=1,\\ So\quad =\quad { (\frac { 9-7\omega }{ 9-7\omega } (\omega )-\frac { 5\omega -3 }{ 5\omega -3 } \times \frac { 1 }{ \omega } ) }^{ 8 }\\ \\ Cancelling\quad like\quad terms,\\ \quad =\quad { (\omega -\frac { 1 }{ \omega } ) }^{ 8 }\\ Putting\quad \omega \quad as\quad \frac { -1+\sqrt { 3 } }{ 2 } ,\\ \quad =\quad { (\frac { -1+\sqrt { 3 } }{ 2 } -\frac { 1 }{ \frac { -1+\sqrt { 3 } }{ 2 } } ) }^{ 8 }\\ Rationalizing\quad the\quad second\quad term,\quad we\quad get\quad \frac { 1 }{ \omega } \quad =\quad { \omega }^{ 2 }\\ \quad =\quad { (\frac { -1+\sqrt { 3 } }{ 2 } -\frac { -1-\sqrt { 3 } }{ 2 } ) }^{ 8 }\\ \quad =\quad { (-\frac { 1 }{ 2 } +\frac { \sqrt { 3 } }{ 2 } +\frac { 1 }{ 2 } +\frac { \sqrt { 3 } }{ 2 } ) }^{ 8 }\\ \quad =\quad { (\sqrt { 3 } })^{ 8 }\quad =\quad 81$

Cheers!:)