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7 answers

I think the closest you could come if you want it to be of the same form is
a^3+b^3+c^3=
(a+b+c) (a^2+b^2+c^2-ab-ac-bc)+3abc

Following this loine of thought i presume the second one would be:

a^3+b^3+c^3+d^3=
(a+b+c+d) (a^2+b^2+c^2+d^2- ab-ac-ad-bc-bd-cd)
+3abc+3abd+3bcd+3acd
=(a+b+c+d) (a^2+b^2+c^2+d^2- ab-ac-ad-bc-bd-cd)
+3abcd(1/a+1/b+1/c+1/c)

2006-08-15 02:39:05 · answer #1 · answered by Anonymous · 1 0

rfamily, you are completely wrong. that is not even a way to factor this expression. moreover, the form that the expression in the question used is totally different than the form that you used.

2006-08-15 02:24:32 · answer #2 · answered by warren g 1 · 0 0

the first grouping is 6, the second is 9
and the third is 12 for a total of 27.

ta-daa!

2006-08-15 01:30:07 · answer #3 · answered by George M 1 · 0 1

I used to know this stuff, but I've forgotten. Look in the beginning of the chapter, it should explain everything.

2006-08-15 01:32:50 · answer #4 · answered by vampire_kitti 6 · 0 0

I don't think that there is any simple way to factor those expressions: at least maxima doesn't know how to do it.

2006-08-15 01:31:45 · answer #5 · answered by Pascal 7 · 0 0

a^3+b^3+c^3=
(a+b+c)^3-3ab(a+b)+3bc(b+c)+3ca(c+a)
a^3+b^3+c^3+d^3
=(a+b)(a^2-ab+b^2)+(c+d)(c^2-cd+d^2)

2006-08-15 01:36:04 · answer #6 · answered by raj 7 · 1 0

where r u from MARS

2006-08-15 01:24:09 · answer #7 · answered by ? 3 · 0 1

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