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⤋