We have,
A+B+C=π......(1)
L.H.S.
cos4A+cos4B+cos4C
2cos(4A+4B)2cos(4A−4B2)+2cos22C−1
=2cos(2A+2B)cos(2A−2B)+2cos22C−1
=2cos2(A+B)cos(2A−2B)+2cos22C−1
=2cos2(A+B)cos(2A−2B)+2cos22C−1
=2cos2(π−C)cos(2A−2B)+2cos22C−1
=2cos2Ccos(2A−2B)+2cos22C−1
=2cos2C[cos(2A−2B)+cos2C]−1
=2cos2C[cos(2A−2B)+cos2(π−(A+B))]−1
=2cos2C[cos(2A−2B)+cos(2A+2B)]−1
=2cos2C[cos(2A+2B)+cos(2A−2B)]−1
=2cos2C[2cos2Acos2B]−1
=4cos2Acos2Bcos2C−1
R.H.S.