Hard to simplify?

Lets try to get the equation in the simpler form. Firstly we will try solving out the " $\log$ " terms in the equation.

$\quad \quad \quad \quad \begin{array}{ll} { 2 }^{ \sqrt { \log _{ 2 }{ 3 } } } & ={ 2 }^{ \cfrac { \log _{ 2 }{ 3 } }{ \sqrt { \log _{ 2 }{ 3 } } } }\\ & ={ 2 }^{ \left( \log _{ 2 }{ 3 } \right) \left( \sqrt { \log _{ 3 }{ 2 } } \right) }\\ & ={ \left( { 2 }^{ \left( \log _{ 2 }{ 3 } \right) } \right) }^{ \sqrt { \log _{ 3 }{ 2 } } }\\ & ={ 3 }^{ \sqrt { \log _{ 3 }{ 2 } } } \end{array}$

$\quad \quad \quad \quad \Rightarrow { 2 }^{ \sqrt { \log _{ 2 }{ 3 } } }-{ 3 }^{ \sqrt { \log _{ 3 }{ 2 } } }=0 .$

And we know the property of logarithms that $a^{\log_a b}=b$ , we can figure out ${ 3 }^{ \log _{ 3 }{ 2 } }-{ 2 }^{ \log _{ 2 }{ 3 } }=2-3=-1$ . So now our equation becomes ${ x }^{ 2 }-3x+2=0$ . Hence using Vieta's Formula we can conclude that $\alpha+\beta=3$ and $\alpha \beta=2$ . So we will solve it further as we move on:

$\quad \quad \quad \begin{array}{ll} { \alpha }^{ 2 }+\alpha \beta +{ \beta }^{ 2 }& ={ \left( \alpha +\beta \right) }^{ 2 }-\alpha \beta \\ & ={ 3 }^{ 2 }-2 \\ & =7 .\square \end{array}$

$(\alpha+\beta )^{2}=\alpha^{2}+2\alpha\beta+\beta^{2}$

$\alpha^{2}+\alpha\beta+\beta^{2} )=(\alpha+\beta)^{2}-\alpha\beta$

$\alpha+\beta=-\frac{B}{A}, \alpha\beta=\frac{C}{A}$

$A=1 , B=-( 3+2^{\sqrt{log_{2} 3}}-3^{ \sqrt{log_{3} ⁡2}}), C=-2(3^{log_{3} ⁡2}-2^{log_{2} ⁡3})$

$(\alpha^{2}+\alpha\beta+\beta^{2} )=(-\frac{B}{A})^{2}-\frac{C}{A}=B^{2}-C$

$=( 3+2^{\sqrt{log_{2} 3}}-3^{ \sqrt{log_{3} ⁡2}})^{2}+2(3^{log_{3} ⁡2}-2^{log_{2} ⁡3})=3^{2}+2(-1)=9-2=7$