2({sin ^6}theta + {cos ^6}theta ) - 3({sin ^4 | vedclass

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Question

(b) ${({\sin ^2}	heta + {\cos ^2}	heta )^3} = {(1)^3}$

$ \Rightarrow \,\,{\sin ^6}	heta + {\cos ^6}	heta + 3\,{\sin ^2}	heta \,{\cos ^2}	heta

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(b) ${({\sin ^2}	heta + {\cos ^2}	heta )^3} = {(1)^3}$

$ \Rightarrow \,\,{\sin ^6}	heta + {\cos ^6}	heta + 3\,{\sin ^2}	heta \,{\cos ^2}	heta = 1$

and ${\sin ^4}	heta + {\cos ^4}	heta + 2\,{\sin ^2}	heta \,{\cos ^2}	heta = 1$

Both gives, $2\,\,({\sin ^6}	heta + {\cos ^6}	heta ) - 3\,({\sin ^4}	heta + {\cos ^4}	heta ) + 1 = 0$.

$2({\sin ^6}\theta + {\cos ^6}\theta ) - 3({\sin ^4}\theta + {\cos ^4}\theta ) + 1 =$

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