Plzzz solve this question and solutionFull solution on the basis iit jee exams i am not able to solve this problem plzzz help me

We are asked to find a^2-2b-2c .Now in equation x^3+ax^2+bx+c -a is sum of root b is product of root taken 2 at a time and -c is product.It given root are cos(€)+cos(¥)+cos(£) where €+£+¥=180 and all three are acute.Now a^2-2b-2c=cos(€)^2+cos(£)^2+cos(¥)^2+2cos(£)cos(¥)+2cos(€).cos(¥)+2cos(€).cos(£)-2cos(£)cos(¥)-2cos(€).cos(¥)-2cos(€).cos(£)+2cos(€).cos(£).cos(¥)=cos(€)^2+cos(£)^2+cos(¥)^2+2cos(€).cos(£).cos(¥)=1 (this is an indentity and i think you can prove it that its value is 1).HENCE ANSWER IS OPTION 2 THAT IS 1.PLEASE APPROVE IF IT HELPS.