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 ___________.

$$ \begin{array}{lllll} \hline x_i & f_i & x_i^2 & f_i x_i & f_i x_i^2 \\ \hline 2 & 4 & 4 & 8 & 16 \\ \hline 4 & 4 & 16 & 16 & 64 \\ \hline 6 & \alpha & 36 & 6 \alpha & 36 \alpha \\ \hline 8 & 15 & 64 & 120 & 960 \\ \hline 10 & 8 & 100 & 80 & 800 \\ \hline 12 & \beta & 144 & 12 \beta & 144 \beta \\ \hline 14 & 4 & 196 & 56 & 784 \\ \hline 16 & 5 & 256 & 80 & 1280 \\ \hline \end{array} $$ <br/><br/>$\therefore$ $\begin{aligned} & \Sigma f_i=40 +\alpha+\beta\end{aligned}$ <br/><br/>$\begin{aligned} & \Sigma f_i x_i=360+ 6 \alpha+12 \beta\end{aligned}$ <br/><br/>$\begin{aligned} & \Sigma f_i x_i^2=3904 +36 \alpha+144 \beta\end{aligned}$ <br/><br/>Given, mean $=9$ <br/><br/>$$ \begin{aligned} & \Rightarrow \frac{\Sigma f_i x_i}{\Sigma f_i}=9 \\\\ & \Rightarrow \frac{360+6 \alpha+12 \beta}{40+\alpha+\beta}=9 \\\\ & \Rightarrow 360+6 \alpha+12 \beta=9(40+\alpha+\beta) \\\\ & \Rightarrow 3 \beta=3 \alpha \\\\ & \Rightarrow \alpha=\beta .........(i) \end{aligned} $$ <br/><br/>$\begin{aligned} & \text { Variance }=15.08 \\\\ & \Rightarrow \frac{\Sigma f_i x_i^2}{\Sigma f_i}-\left(\frac{\Sigma f_i x_i}{\Sigma f_i}\right)^2=15.08 \\\\ & \Rightarrow \frac{3904+36 \alpha+144 \beta}{40+\alpha+\beta}-(9)^2=15.08 \\\\ & \Rightarrow \frac{3904+180 \alpha}{40+2 \alpha}=81+15.08 \quad[\because \alpha=\beta] \\\\ & \Rightarrow 3904+180 \alpha=96.08(40+2 \alpha) \\\\ & \Rightarrow 3904+180 \alpha=3843.2+192.16 \alpha \\\\ & \Rightarrow 60.8=12.16 \alpha \\\\ & \Rightarrow \alpha=5=\beta \\\\ & \therefore \alpha^2+\beta^2-\alpha \beta=25+25-25=25\end{aligned}$