question_answer

If a and b are the roots of the equation \[{{x}^{2}}-4x+1=0\] \[(a>b)\]then the value of \[f(\alpha ,\beta )=\frac{{{\beta }^{3}}}{2}\text{cose}{{\text{c}}^{2}}\left( \frac{1}{2}{{\tan }^{-1}}\frac{\beta }{\alpha } \right)+\frac{{{\alpha }^{3}}}{2}{{\sec }^{2}}\left( \frac{1}{2}{{\tan }^{-1}}\frac{\alpha }{\beta } \right)\]is

Answer:

56