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Home » I = \frac(n-1)(n-2)2 - Aurero

I = \frac(n-1)(n-2)2 - Aurero

📅 March 13, 2026 👤 scraface

Mar 13, 2026

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Question: Factor the expression $ 16x^4 - 81y^4 $.
Recognize this as a difference of squares:
16x^4 - 81y^4 = (4x^2)^2 - (9y^2)^2 = (4x^2 - 9y^2)(4x^2 + 9y^2)
Now notice that $ 4x^2 - 9y^2 $ is also a difference of squares:
4x^2 - 9y^2 = (2x - 3y)(2x + 3y)
So the full factorization is:
Final answer: $ \boxed{(2x - 3y)(2x + 3y)(4x^2 + 9y^2)} $
Question: Compute the square of $ 3\sqrt{2} - 2\sqrt{3} $.
Use the identity $ (a - b)^2 = a^2 - 2ab + b^2 $:
Compute each term:

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