Question

Sin⁸Q-Cos⁸ Q =(sin²Q-cos²Q)(1-2Sin2Q Cos Q)​

Answer 1

Find the above attached as you requested.

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Answer 2

$\sf{\underline{Appropriate \: question:}}$

Prove that :

$(sin^{8} θ - cos^{8} θ) = (sin^{2} θ - cos^{2} θ)(1 - 2sin^{2} θ cos^{2} θ)$

$\sf{\underline{Solution:}}$

We have,

$(sin⁸θ - cos⁸θ) = (sin⁴θ)² - (cos⁴θ)² = (sin⁴θ - cos⁴θ)(sin⁴θ + cos⁴θ)$

$⇒ LHS = (sin²θ - cos²θ)(sin²θ + cos²θ) (sin⁴θ + cos⁴θ)$

$⇒ LHS = (sin²θ - cos²θ) [(sin²θ)² + (cos²θ)² + 2sin²θcos²θ - 2sin²θcos²θ]$

$⇒ LHS = (sin²θ - cos²θ) [(sin²θ) + (cos²θ)²</p><p> - 2sin²θcos²θ]$

$⇒LHS = (sin²θ - cos²θ)(1 - 2sin²θcos²θ) = RHS$

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