So basically I have the answer to this problem because it's in the book, but I have no clue how the author solved it. It's from Schaum's Precalculus. Thanks.
Factor $(16x^4)-(x^2y^2)+(y^4) = (4x^2+y^2-3xy)(4x^2+y^2+3xy)$
2026-04-08 02:29:55.1775615395
Factor $16x^4-x^2y^2+y^4$
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Hint
$$16x^4+y^4=(4x^2+y^2)^2-8x^2y^2$$ $$16x^4-x^2y^2+y^4=(4x^2+y^2)^2-9x^2y^2=(4x^2+y^2)^2-(3xy)^2$$ Now, use $a^2-b^2=(a+b)(a-b)$