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factor completely m^6-216n^6 i want to know if i did it right..thanks
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i get (m^2-6n^2)(m^4+6n^2m^2+36n^4)
you are correct
That looks correct to me lordofpens. :)
This is a difference of cubes,
ty
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how about 121a^2-(3b^3-5x)^2
\[m^6 -216n^6 = (m^2)^3-(6n^2)^3\]\[a^3-b^3 = (a-b)(a^2+ab+b^2)\]
i get (11a-3b63+5x)(11a+3b^3-5x)
Idk what you're doing there, but your first answer was correct :)
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