Hey guys, I have some math homework...
The last equation I can't solve is that:
(x^2+x+1)^2(x^2-x+1)=3
Please help me solve that one
Thanks!
Expanded out your equation is:Originally Posted by dgolverk
$\displaystyle x^6+x^5+2x^4+x^3+2x^2+x-2=0$
Using the rational roots test the possible rational zeros are x = +/- 1, +/- 2. The only rational solution is x = -1. Using synthetic division on the zero x = -1 gives:
$\displaystyle x^5+2x^3-x^2+3x-2=0$
which does have a real solution near x = 0.611 (so the other 4 are complex), but being a 5th degree polynomial equation I don't have a way to generally solve it exactly.
-Dan
There may be a trick involved. I noted when I looked at the problem that each factor is related to the factorization of the sum and difference of two cubes:Originally Posted by dgolverk
$\displaystyle x^3 - 1 = (x-1)(x^2+x+1)$
and
$\displaystyle x^3+1=(x+1)(x^2-x+1)$
However, I can't think of a way to use this information. That 3 on the RHS spoils things pretty badly for me. Sorry!
-Dan