August899
  • August899
Factor the polynomials:m^3+1/8 and p^3+216
Algebra
katieb
  • katieb
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zepdrix
  • zepdrix
Hey :)
zepdrix
  • zepdrix
\[\large\rm m^3+\frac{1}{8}\]Hmmm, recall that we have a rule for factoring the `sum of cubes`: \[\large\rm a^3+b^2=(a+b)(a^2-ab+b^2)\]
zepdrix
  • zepdrix
But with the way our problem is written right now, we only have one cube. Any ideas how to turn the 1/8 into `something`^3 ?

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zepdrix
  • zepdrix
August899
  • August899
do we divide 1 over 8?
zepdrix
  • zepdrix
The 8 is some number to the third power.
zepdrix
  • zepdrix
Example: 27 is 3 to the third power. Because 3 x 3 x 3 = 27 Therefore \(\large\rm 27=3^3\)
zepdrix
  • zepdrix
Comeon Auguuuuuust >.< break down the 8, how can we write that another way?

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