(6 sin^3 β + 6 cos^3 β)/ (sin β + cos β) = 6 - 6sin β + cos β

Question

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Answer 1

Given identity :

(6 sin³β + 6 cos³β) / (sinβ +cosβ) = 6 - 6sinβ cosβ

Start from left hand side

(6 sin³β + 6 cos³β) / (sinβ +cosβ)

6 (sin³β + cos³β) / (sinβ +cosβ)

Substitute formula : a³ + b³ = (a + b)(a² - ab + b²)

6 [ (sinβ +cosβ)(sinβ² - sinβcosβ + cos²β) ] / (sinβ +cosβ)

6 [ (sinβ² - sinβcosβ + cos²β) ]

6 (sinβ² + cos²β - sinβcosβ )

Substitute formula : sinβ² + cos²β = 1

6 (1 - sinβcosβ )

6  - 6sinβcosβ

Given identity (6 sin³β + 6 cos³β) / (sinβ +cosβ) = 6 - 6sinβ cosβ is proved.