Johnny Kwong

Posted on Monday, 05 May, 2003 - 12:35 am:

Does anyone know how to factorise x⁶ +2 over the reals? I know it has no real roots so it must factorise into quadratics, but I got nowhere.

Please help

Michael DorÃ?©

Posted on Monday, 05 May, 2003 - 01:02 am:

Hint: first find its complex roots, then pair up these roots into conjugate pairs.

Demetres Christofides

Posted on Monday, 05 May, 2003 - 09:09 am:

Further hint: 6th roots of unity.

Demetres

Demetres Christofides

Posted on Tuesday, 06 May, 2003 - 08:53 am:

Maybe that was not helpful enough. The factorisation of

x^{6} - 1

(x - 1) (x - ω) (x - ω^{2}) (x - ω^{3}) (x - ω^{4}) (x - ω^{5})

where

ω = \exp (2 π i / 6)

Can you now factorise $x^{6} + 2$ over the complex numbers? Demetres