I'm having a tough time understanding these:

X^6 - 1

http://img.nattawat.org/images/kqb6m...6sd4_thumb.jpg

http://img.nattawat.org/images/c0ix0...tpwz_thumb.jpg

And if you can, please give me a little explanation on how you got the answer.

- Jan 12th 2011, 02:19 PMConnor[Solved]Having trouble w/ a few Factoring Polynomials and Radical Expression problems
I'm having a tough time understanding these:

X^6 - 1

http://img.nattawat.org/images/kqb6m...6sd4_thumb.jpg

http://img.nattawat.org/images/c0ix0...tpwz_thumb.jpg

And if you can, please give me a little explanation on how you got the answer. - Jan 12th 2011, 02:21 PMdwsmith
$\displaystyle 48=3*16$

$\displaystyle 75=3*25$

$\displaystyle \displaystyle\frac{5\sqrt{16}\sqrt{3}+\sqrt{25}\sq rt{3}}{5\sqrt{3}}$

Does this help for the first one? - Jan 12th 2011, 02:22 PMdwsmith
$\displaystyle x^6-1=0\Rightarrow x^6=1 \ \ \ x=\pm 1$

Do you want all the solutions or just the two real solutions? - Jan 12th 2011, 02:24 PMdwsmith
$\displaystyle (4\sqrt{2}-3\sqrt{5})2\sqrt{25}=4\sqrt{2}*2\sqrt{25}-3\sqrt{5}*2\sqrt{25}=\cdots$

- Jan 12th 2011, 02:27 PMpickslides
- Jan 12th 2011, 02:30 PMConnor
- Jan 12th 2011, 02:34 PMConnor
- Jan 12th 2011, 02:34 PMdwsmith
- Jan 12th 2011, 02:35 PMdwsmith
- Jan 12th 2011, 02:36 PMe^(i*pi)
Once you've used the difference of two squares (as Pickslides suggests) then you can use the sum/difference of two cubes

$\displaystyle (a \pm b)^3 = (a \pm b)(a^2 \mp ab + b^2)$

Since x^6 and 1 are both square numbers and cube numbers you can start with either formula - Jan 12th 2011, 02:41 PMConnor
- Jan 12th 2011, 02:44 PMdwsmith
$\displaystyle \displaystyle\frac{5\sqrt{16}\sqrt{3}+\sqrt{25}\sq rt{3}}{5\sqrt{3}}\Rightarrow \left(\frac{\sqrt{3}}{\sqrt{3}}\right)\left(\frac{ 5\sqrt{16}+\sqrt{25}}{5}\right)\Rightarrow\frac{5\ sqrt{16}+5}{5}\Rightarrow\left(\frac{5}{5}\right)( \sqrt{16}+1)=4+1=5$

It should probably really be $\displaystyle \pm 5$ but you may only need the positive solution. - Jan 12th 2011, 02:46 PMConnor
- Jan 12th 2011, 02:48 PMdwsmith
Do the last problem in the same method. I have already completed one step for you, see post 4.

- Jan 12th 2011, 02:55 PMConnor