The answer in itself is 1/128 and here is the procedure to prove it:
cos(A)*cos(60+A)*cos(60-A) = cos(A)*(cos²60 - sin²A)
= cos(A)*{(1/4) - 1 + cos²A} = cos(A)*(cos²A - 3/4)
= (1/4){4cos^3(A) - 3cos(A)} = (1/4)*cos(3A)
Now we group applying what we see above
cos(12)*cos(48)*cos(72) =
=cos(12)*cos(60-12)*cos(60+12) = (1/4)cos(36)
Similarly, cos(24)*cos(36)*cos(84) = (1/4)cos(72)
Now the given expression is:
= (1/4)cos(36)*(1/4)*cos(72)*cos(60) =
= (1/16)*(1/2)*{(√5 + 1)/4}*{(√5 - 1)/4} [cos(60) = 1/2;
cos(36) = (√5 + 1)/4 and cos(72) = cos(90-18) =
= sin(18) = (√5 - 1)/4]
And we seimplify it and it goes: (1/512)*(5-1) = 1/128
