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

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

Sin⁶ + Cos⁶ = 1 - 3Sin^​2Cos²
Sin⁶ + Cos⁶ + 3Sin^​2Cos² =  1
(Sin²)³ + (Cos²)³ + 3Sin^​2Cos^{2 = 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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Ans1

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