Question -
Answer -
Let us consider LHS: sin6┬аx+ cos6┬аx
(sin2┬аx)┬а3┬а+(cos2┬аx)┬а3
By using the formula, a3┬а+ b3┬а=(a + b) (a2┬а+ b2┬атАУ ab)
(sin2┬аx + cos2┬аx)[(sin2┬аx)┬а2┬а+(cos2┬аx)┬а2┬атАУsin2┬аx cos2┬аx]
By using the formula, sin2┬аx + cos2┬аx= 1 and a2┬а+ b2┬а= (a + b)┬а2┬атАУ2ab
1 ├Ч [(sin2┬аx + cos2┬аx)┬а2┬атАУ 2sin2┬аxcos2┬аx тАУ sin2┬аx cos2┬аx
12┬атАУ3sin2┬аx cos2┬аx
1 тАУ 3sin2┬аx cos2┬аx
= RHS
тИ┤LHS = RHS
Hence proved.