If f(x)=3[sin4(3π4−x)+sin4(3π+x)]−2[sin6(π2+x)+sin6(5π−x)] then, for all permissible values of x, f(x) is -
3[sin4(3π2−α)+sin4(3π+α)] - [sin6(π2+α)+sin6(5π−α)] =
The expression 3[sin4(3π2−α)+sin4(3π+α)]−2[sin6(π2+α)+sin6(5π+α)] is equal to