\operatorname { sin } ^ { 3 } x + \operatorname { cos } ^ { 3 } x = (

1.	\operatorname { sin } ^ { 3 } x + \operatorname { cos } ^ { 3 } x = ( \operatorname { sin } x + \operatorname { cos } x ) ( 1 - \operatorname { sin } x \operatorname { cos } x )
Answer» we can use the identity a³+b³ = (a+b)(a²+b²-ab) here a = sinx , b=cosx so sin³x+ cos³x = (sinx+cosx)(sin²x+cos²x-sinxcosx).but sin²x + cos²x = 1 so, sin³x+ cos³x = (sinx+cosx)(1-sinxcosx).

\operatorname { sin } ^ { 3 } x + \operatorname { cos } ^ { 3 } x = ( \operatorname { sin } x + \operatorname { cos } x ) ( 1 - \operatorname { sin } x \operatorname { cos } x )

Answer»

we can use the identity a³+b³ = (a+b)(a²+b²-ab)

here a = sinx , b=cosx so sin³x+ cos³x = (sinx+cosx)*(sin²x+cos²x-sinxcosx).but sin²x + cos²x = 1 so, sin³x+ cos³x = (sinx+cosx)*(1-sinxcosx).