Prove that:- sin6 theta + cos6 theta = 1-3sin2theta.cos2theta

 sin^6 x + cos^6 x

= (sin^2 x + cos^2 x) ( sin^4 x - sin^2 cos^2 x + cos^4 x)

= (sin^4 x + cos^4 x) - sin^2 x cos^2 x

= (sin^2 x + cos^2 x)^2 - 2 sin^2 cos^2 x - sin^2 x cos^2 x

= 1 - 3 sin^2 x cos^2 x

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Consider, sin6ө + cos6ө =1-3sin2өcos2ө
(sin2ө)3 + (cos2ө)3 = (sin2ө + cos2ө)3 − 3 (sin2ө cos2ө)(sin2ө + cos2ө)   [since a + b = (a+b)3 − 3ab(a+b)]
                           = 1 − 3sin2өcos2ө    [Since sin2ө + cos2ө = 1] 
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cos6 - cos10
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Sin6 + Cos6 = 1 - 3Sin​2Cos2
Sin+ Cos6 + 3Sin​2Cos2 =  1
(Sin2)3 + (Cos2)3 + 3Sin​2Cos2 = 1

[a3 + b3 = (a+b)(a2 + b2 - ab)] 

(Sin2 + Cos2)(Sin4 + Cos4 - Sin2Cos2)  + 3Sin​2Cos2

[Sin2 + Cos2 = 1]

(1)(Sin4 + Cos4 - Sin2Cos2) ​+ 3Sin​2Cos2

Sin4 + Cos4 - Sin2Cos2 ​+ 3Sin​2Cos2

Sin4 + Cos4 + 2Sin2Cos2 ​    

[a2 + b2 + 2ab = (a + b)2]

(Sin2 + Cos2)2 

(1)2 

= 1 = RHS.

 
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sin^6 x + cos^6 x = (sin^2 x + cos^2 x) ( sin^4 x - sin^2 cos^2 x + cos^4 x) = (sin^4 x + cos^4 x) - sin^2 x cos^2 x = (sin^2 x + cos^2 x)^2 - 2 sin^2 cos^2 x - sin^2 x cos^2 x = 1 - 3 sin^2 x cos^2 x
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