The problem of decomposing of the given expression is well known, as well as its sophisticated solutions. Now, the problem is to find an EASY solution!
a3+b3+c3-3abc=(a+b)3-3a2b-3ab2+c3-3abc
=(a+b+c)3-3(a+b)2c-3(a+b)c2-3ab(a+b+c)
=(a+b+c)3-3(a+b)c(a+b+c)-3ab(a+b+c)=(a+b+c)(a2+b2+c2-ab-bc-ac)
Yes, Ioulia! Congratulations! I was almost sure that you will be the first who will give the correct solution.
(By the way, we are both from MGPI!)
I think it is interesting:(if I am right)
a3 + b3 = (a + b)(a2 + b2 -ab)
a3 + b3 + c3 = (a + b + c)(a2 + b2 + c2 - ab - ac - bc)
STOP (the assumption is the same;: real coefficients)
a3 + b3 + c3 + d3 = (a + b + c + d)(?...?)
Greetings, Tadeusz
I am very surprised that for this question I was given downvote. 17 views, 2 upvotes for its solution and a downvote for the question... How may it be? It is the second time in my not so long history here. Who are doing that? Do they understand what are they doing?
Added March 9, 2016: 89 views, 8 upvotes for the solution which I have in mind when asked this question, and ... downvote for the question itself! Does it not bother the admins of RG?
While considering this question; quite unfortunately not from the simple perspective requested, I have found that P = a3 + b3 + c3 - 3abc = 2S, where S is the area of the triangle of vertices: (a2 - bc, c), (b2 - ac, a), (c2 - ab, b). The link given below (*) recalls how can be obtained as S = (1/2)·D, where D is the (3x3) determinant given as ― 1st row: |(a2 - bc) c 1|, 2nd row: |(b2 - ac) a 1|, 3rd row: |(c2 - ab) b 1|. This determinant corresponds to the given polynomial expression: D = (a2 - bc)·(a - b) + (b2 - ac)·(b - c) + (c2 - ab)·(c - a) =(a2 - bc)·a + (b2 - ac)·b + (c2 - ab)·c = P.
(*) Cf. eq. (16): http://mathworld.wolfram.com/TriangleArea.html
Please allow me to complete the reasoning proposed with above post ― Let us further consider the triangle of vertices: v1 ≡ (a2 - bc, c), v2 ≡ (b2 - ac, a), v3 ≡ (c2 - ab, b), proposed above. The lengths of the triangle sides would be given by: d12 = √{[(a2 - bc) - (b2 - ac)]2 + (c-a)2}, d23 = √{[(b2 - ac) - (c2 - ab)]2 + (a-b)2}, d13 = √{[(a2 - bc) - (c2 - ab)]2 + (c - b)2}. With the semiperimeter given by p = (1/2)·(d12 + d23 + d13), the triangle area can be given by Heron’s formula: S = √[p·(p - d12)·(p - d23)·(p – d13)].
- Badges
- Science topic
- Similar topics
- Psychology