1. ## 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

2. 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

3. Hello,

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

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