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View Full Version : Roots of x^3 + 216


Alkatran
Sep18-04, 01:59 PM
I need to factorise x^3 + 216 to include (x+6):

(x + 6) \ (x^3 + 216) (lim where x approaches -6)

I broke it down to (x + 6i)(x^2 - 36i) ... but that's no good.

arildno
Sep18-04, 02:05 PM
Well, shouldn't polynomial division work fine?

Pyrrhus
Sep18-04, 02:07 PM
Remember

X^3 + Y^3 = (X+Y)(X^2-XY+Y^2)

Pyrrhus
Sep18-04, 02:08 PM
He has

\frac{x+6}{x^3 + 216}

\frac{x+6}{x^3 + 6^3}

Alkatran
Sep18-04, 06:31 PM
I got it fine before I even came back. I was trying to remember how to do long division and the cube rule that you guys posted. The start of the year, still getting restarted...