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Sums and differences of cubes??

Factor Completly:

1) 3z^5 - 3z^2

2) a^9 + b^12 c^15

2007-09-05 00:04:01 · 2 answers · asked by Anonymous in Science & Mathematics Mathematics

2 answers

1) 3z^2 (z-1) (z^2 + z +1)

2) (a^3 + b^4c^5) (a^6 - a^3b^4c^5 + b^8c^10)

2007-09-05 00:15:45 · answer #1 · answered by furball 4 · 0 0

3z^5 - 3z^2 = 3z^2(z^3 - 1)

a^9 + b^12 c^15 = a^9 + (bc)^12 *c^3 ? what else can be done ? (a^3)^3 + ((bc)^3)^4 *c^3

2007-09-05 07:21:59 · answer #2 · answered by Beardo 7 · 0 0

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