how do you factorize x^4+27x

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how do you factorize x^4+27x

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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Hey Angela! They've both got some x's... factor those out first.\[\large\rm x^4+27x=x(x^3+27)\]From there you want to utilize your `Sum of Cubes Formula`: \(\large\rm a^3+b^3=(a+b)(a^2-ab+b^2)\)
Hmm but we only have one thing being cubed... x. If you can find a way to rewrite 27 as a perfect cube, you can apply this formula!
but how do you get (a+b)(a^2-ab+b^2) from x(x^3+27)? @zepdrix

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Other answers:

You get \(\large\rm (a+b)(a^2-ab+b^2)\) from \(\large\rm a^3+b^3\) and you get \(\large\rm a^3+b^3\) from \(\large\rm x^3+27\)
the x^3 matches up with the a^3, it's in the correct format. But we need to rewrite 27 as \(\large\rm something^3\)
Example: \(\large\rm 8=2\cdot2\cdot2\) therefore \(\large\rm 8=2^3\) We found a way to write 8 as a perfect cube! Can you find a way to do that with 27?
thank you!

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