Your matrix is the product of AB.
Assuming that det (AB)= det(A)det(B) for 3 x 3 matrices, prove that
| 3 a+b+c a^3+b^3+c^3 |
| a+b+c a^2+b^2+c^2 a^4+b^4+c^4 |
| a^2+b^2+c^2 a^3+b^3+c^3 a^5+b^5+c^5 |
is equal to (a+b+c) (a-b)^2 (a-c)^2 (b-c)^2
I do this and get 0 which doesnt help me at all
Hello, feage7!
Who assigned this problem . . . Professor deSade?
Assuming that prove that
Thanks to Plato's incredible factoring, the rest is "just algebra" . . . Ha!
It still takes Olympic-level gymnastics to simplify the determinants.
It involves at lot of "factor by grouping".
I won't bother explaining every step; I hope you can follow anyway.
We have: /
. .
. .
. .
. .
Hence: .
We have: .
. .
. .
. .
. .
. .
. .
. .
Hence: .
Therefore: .
. . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . ta-DAA!
I need a nap . . .
yeah i know that i mean the question asks for (a+b+c)(a-b)^2(a-c)^2(b-c)^2 not (a+b+c)(a-b)^2(c-a)^2(b-c)^2 which is why im saying u swap (a-c) and (c-b) to get the (b-c) term correct but then (a-c) becomes (c-a) in the final answer which means by making one term right ur making another wrong.
You want and have . The only difference is between and . But:
so there is no difference.
With respect to each determinant, Both of Soroban's Det(A) and Det(B) had a single factor of (c-a) so switching to (a-c) in both gives a negative sign for each. But since we are multiplying the two determinants, we get two negative signs, which cancel.
-Dan