Question

hi solve this question
use a^3+b^3= (a+b)(a^2-ab+b^2) formula

Answer 1

Answer:

QED

Step-by-step explanation:

Sin^6θ + cos^6θ  + 3Sin^2θCos^2θ

= (sin^2θ+Cos^2θ)(Sin^4θ-Sin^2θCos^2θ+Cos^4θ) + 3sin^2θCos^2θ

Using a^3+b^3= (a+b)(a^2-ab+b^2)  a = Sin^2θ   b = Cos^2θ

as sin^2θ + Cos^2θ = 1

=  Sin^4θ +2sin^2θCos^2θ+Cos^4θ

= (sin^2θ + Cos^2θ)^2

= 1^2

= 1

QED

= 1

Answer 2

Answer:

1

Step-by-step explanation:

Given Equation is sin⁶θ + cos⁶θ + 3sin²θcos²θ

= (sin²θ)³ + (cos²θ)³ + 3sin²θcos²θ

= (sin²θ + cos²θ)(sin⁴ + cos⁴θ - sin²θcos²θ) + 3sin²θcos²θ

= (sin²θ + cos²θ)[sin⁴θ + cos⁴θ - sin²θcos²θ + 3sin²θcos²θ]

= (1)[sin⁴θ + cos⁴θ + 2sin²cos²θ]

= (sin²θ)² + (cos²θ)² + 2(sin²θ)(cos²θ)

= (sin²θ + cos²θ)²

= 1.

Hope it helps!