# simplifying this equation

• May 6th 2008, 08:58 AM
thermalwarrior
simplifying this equation
how does this:

(3 - &)(2-&)(4-&) - 2(8-2&)-2(4-2&)= 0

where & is equal to lambada(sorry cant seem to get symbol on)

end up as this:

&^3 - 9&^2 + 18& = 0

cheers
• May 6th 2008, 01:01 PM
topsquark
Quote:

Originally Posted by thermalwarrior
how does this:

(3 - &)(2-&)(4-&) - 2(8-2&)-2(4-2&)= 0

where & is equal to lambada(sorry cant seem to get symbol on)

end up as this:

&^3 - 9&^2 + 18& = 0

cheers

Small wonder that you are having troubles: It doesn't.

Your expression comes out as $\displaystyle -\lambda ^3 + \lambda ^2 + 22 \lambda - 40$

-Dan
• May 6th 2008, 01:42 PM
Moo
Hello,

Quote:

Originally Posted by topsquark
Small wonder that you are having troubles: It doesn't.

Your expression comes out as $\displaystyle -\lambda ^3 + \lambda ^2 + 22 \lambda - 40$

-Dan

I find it to be $\displaystyle -\lambda^3+9 \lambda^2-18 \lambda=0$, simply by developping (Surprised)

This equation is the one given (divide by -1).