Prove that the polynomial $$x^{3n}+x^{3m+1}+x^{3k+2}$$ is exactly divisible by $$x^2+x+1$$ if m, n, k are any non-negative integers.

I know only this much that $$x^2+x+1$$
becomes zero at $$\omega \space and \space \omega ^2$$ .

Question

Prove that the polynomial $$x^{3n}+x^{3m+1}+x^{3k+2}$$ is exactly divisible by $$x^2+x+1$$ if m, n, k are any non-negative integers.

I know only this much that  $$x^2+x+1$$
becomes zero at $$\omega \space and \space \omega ^2$$ .

maheshmeghwal9 · Answer

@waterineyes Plz help:)

anonymous · Answer

Frankly speaking I have no clue how to do this but I am thinking of it..

maheshmeghwal9 · Answer

ok np :)
but i have done something

anonymous · Answer

Yes you can show that to me..

anonymous · Answer

Can anyone here help us...

anonymous · Answer

we see that by substuting omega in the first equation the result is zer0......similarly with omega squared its the same result.......got any idea???

maheshmeghwal9 · Answer

I have done
put omega in $$1+x+x^2$$$$1+ \omega + \omega ^2=0$$so put x = omega in $$\omega^{3n}+\omega^{3m+1}+\omega^{3k+2}=1+\omega +\omega^2=0$$

anonymous · Answer

Now try with w^2 as Prophet is saying..

anonymous · Answer

yup.....we get zero even by substituting omega squared........

that means .....by factor theorem first equation can be written as
$$(x-\omega)(x-\omega^{2})(some polymonial)$$

maheshmeghwal9 · Answer

i m getting same with omega squred

anonymous · Answer

and second equation is 
$$(x-\omega)(x-\omega^{2})$$

maheshmeghwal9 · Answer

but we have to prove:)

maheshmeghwal9 · Answer

how can we prove tht statement?

anonymous · Answer

divide them both......by cancelling 
$$(x-\omega)(x-\omega^{2})$$

we get some polymonial ......
hence proved!!!!

maheshmeghwal9 · Answer

so both equation can be written as $$(x- \omega)(x-\omega^2)$$therefore
$$\frac {\cancel{(x-\omega)(x-\omega^2)}}{\cancel{(x-\omega)(x-\omega^2)}}=1.$$hence proved.
Am i right?

anonymous · Answer

no!!!!!

the first equation can be written as 
(x-w)(x-w^2)*(some polynomial)...........

its same as in wht is 24 divisible by 3.......
its because 3 is a factor of 24.....
24=3*12

$$\frac{24}{3}=\frac{3*12}{3}$$

maheshmeghwal9 · Answer

but the first eq. is also like this $$1+\omega+\omega^2$$then it must be only
$$(x-\omega)(x-\omega^2).$$

anonymous · Answer

by substituting and getting zero we mean that omea and omega squared are factors of the first equation......just google 'factor theorem' .....u'll know what i mean...

maheshmeghwal9 · Answer

ohh i seeeeeeeeeee
thanx a lot
now i gt wt u r saying:)

anonymous · Answer

pleasure!!!