If the mean and variance of the frequency distribution

$x_i$	2	4	6	8	10	12	14	16
$f_i$	4	4	$\alpha$	15	8	$\beta$	4	5

are 9 and 15.08 respectively, then the value of  ${\alpha }^2+{\beta }^2+\alpha \beta$  is_______

Given, Mean and variance are 9 and 15.08 respectively

Mean  $=\frac{8+16+120+80+56+80+6\alpha +12\beta }{40+\alpha +\beta }=9$
 $\Rightarrow 360+9\alpha +9\beta =360+6\alpha +12\beta$
 $\Rightarrow 3\alpha -3\beta =0$
 $\Rightarrow \alpha =\beta \mathrm{........}\left(1\right)$
Now variance is given by,
 $15.08=\frac{16+64+36\alpha +960+800+144\alpha +784+1280}{40+2\alpha }-9^2$
 $\Rightarrow 96.08=\frac{3904+180\alpha }{40+2\alpha }$
 $\Rightarrow (40+2\alpha )\left(96.08\right)=3904+180\alpha$
 $\Rightarrow 384.3+\left(192.16\right)\alpha =3904+180\alpha$
 $\Rightarrow \left(12.16\right)\alpha =60.80$
 $\Rightarrow \alpha =5=\beta$
Hence,  ${\alpha }^2+{\beta }^2+\alpha \beta =25+25+25=75$