There's another way to solve this?

[Vali]

There is x a complex number.
$\displaystyle x^{3}=1$ so $\displaystyle (x-1)(x^{2}+x+1)=0$
I need to find the value of $\displaystyle (1+x)(1+x^{2})(1+x^{3})(1+x^{4})(1+x^{5})(1+x^{6})$
I solved the exercise but I didn't have an algorithm to follow.I just "played" with operations until I got something. (4 final answer which is correct)
I combined the first and the second terms and I multiplied them and I got 2(1+x^4)(1+x^5)(1+x^6)
The last term is 1+x^3*x^3 = 2
So I have 4(1+x^4)(1+x^5)
I multiplied the last two terms and I got 4(1+x^5+x^4+x^3*x^3) = 8 + 4(x^5+x^4) = 8 + 4x^4(x+1)
That "x+1" is -x^2 from x^3=1 so I got 8 - 4x^3 = 8-4 =4
My question is if there's another way to solve this exercise.I don't know, maybe it's something which I should notice.Something tricky, an "elegant" method.
Thanks!

 

[pka]

There is x a complex number.
$\displaystyle x^{3}=1$ so $\displaystyle (x-1)(x^{2}+x+1)=0$
I need to find the value of $\displaystyle (1+x)(1+x^{2})(1+x^{3})(1+x^{4})(1+x^{5})(1+x^{6})$
I solved the exercise but I didn't have an algorithm to follow.I just "played" with operations until I got something. (4 final answer which is correct)
I combined the first and the second terms and I multiplied them and I got 2(1+x^4)(1+x^5)(1+x^6)
The last term is 1+x^3*x^3 = 2
So I have 4(1+x^4)(1+x^5)
I multiplied the last two terms and I got 4(1+x^5+x^4+x^3*x^3) = 8 + 4(x^5+x^4) = 8 + 4x^4(x+1)
That "x+1" is -x^2 from x^3=1 so I got 8 - 4x^3 = 8-4 =4
My question is if there's another way to solve this exercise.I don't know, maybe it's something which I should notice.Something tricky, an "elegant" method.

This is another example where we need the exact wording of the question.
There are three numbers that satisfy $\displaystyle x^{3}=1$: $\displaystyle \exp \left( {\frac{{k2\pi i}}{3}} \right),~k=0,1,2$
See here for $\displaystyle k=1$.
Please tell us the exact wording of the question.

 

[Dr.Peterson]

There is x a complex number.
$\displaystyle x^{3}=1$ so $\displaystyle (x-1)(x^{2}+x+1)=0$
I need to find the value of $\displaystyle (1+x)(1+x^{2})(1+x^{3})(1+x^{4})(1+x^{5})(1+x^{6})$
I solved the exercise but I didn't have an algorithm to follow.I just "played" with operations until I got something. (4 final answer which is correct)
I combined the first and the second terms and I multiplied them and I got 2(1+x^4)(1+x^5)(1+x^6)
The last term is 1+x^3*x^3 = 2
So I have 4(1+x^4)(1+x^5)
I multiplied the last two terms and I got 4(1+x^5+x^4+x^3*x^3) = 8 + 4(x^5+x^4) = 8 + 4x^4(x+1)
That "x+1" is -x^2 from x^3=1 so I got 8 - 4x^3 = 8-4 =4
My question is if there's another way to solve this exercise.I don't know, maybe it's something which I should notice.Something tricky, an "elegant" method.
Thanks!

First, I suppose when you say x is complex, you mean it is one of the non-real roots, so that $\displaystyle x^{2}+x+1=0$.

Using some of the same ideas you found, I would first note that since $\displaystyle x^{3}=1$, $\displaystyle x^{k+3}=x^k$. So we can immediately simplify the expression you want as $\displaystyle (1+x)(1+x^{2})(1+1)(1+x)(1+x^{2})(1+1) = 4((1+x)(1+x^2))^2$.

You can take it from there, using facts you've already stated.

 

[Vali]

Thank you for your help.

 

[JeffM]

I am confused. What does it mean to combine the first two factors and then multiply them?

As tkhunny says, correct answers do not care how you find them. Exploring is a perfectly good algorithm.

I must admit what I would try is this, but it depends on 'seeing" something.

First off [MATH]x^3 = 1 \implies x = 1, \ x = \dfrac{-\ 1 + i\sqrt{3}}{2}, \text { or } x = \dfrac{-\ 1 - i\sqrt{3}}{2}.[/MATH]
Let's check.

[MATH]x = \dfrac{-\ 1 + i\sqrt{3}}{2} \implies x^2 = \dfrac{1 - 2i\sqrt{3} - 3}{4} = \dfrac{- \ 2 - 2i\sqrt{3}}{4} = \dfrac{-\ 1 - i\sqrt{3}}{2} \implies[/MATH]
[MATH]x^3 = x * x^2 = \dfrac{-\ 1 + i\sqrt{3}}{2} * \dfrac{-\ 1 - i\sqrt{3}}{2} = \dfrac{1 - i^2 * 3}{4} = \dfrac{1 + 3}{4} = 1.[/MATH]
[MATH]x = \dfrac{-\ 1 - i\sqrt{3}}{2} \implies x^2 = \dfrac{1 + 2i\sqrt{3} - 3}{4} = \dfrac{- \ 2 + 2i\sqrt{3}}{4} = \dfrac{-\ 1 + i\sqrt{3}}{2} \implies[/MATH]
[MATH]x^3 = x * x^2 = \dfrac{-\ 1 - i\sqrt{3}}{2} * \dfrac{-\ 1 + i\sqrt{3}}{2} = \dfrac{1 - i^2 * 3}{4} = \dfrac{1 + 3}{4} = 1.[/MATH]
OK, that all checks. So now

[MATH]x = 1 \implies \prod_{k=1}^6 (1 + x^k) = \prod_{k=1}^6 (1 + 1^k) = 2^6 = 64. [/MATH]
But what if x does not = 1.

[MATH]x^3 = 1 \implies (x^3)^2 = 1^2 = 1 \implies (1 + x^3)(1 + x^6) = (1 + 1)(1 + 1) = 4.[/MATH]
[MATH]x^4 = x^3 * x = 1 * x = x \implies (1 + x)(1 + x^4) = (1 + x)^2.[/MATH]
[MATH]x^5 = x^3 * x^2 = 1 * x^2 = x^2 \implies (1 + x^2)(1 + x^5) = (1 + x^2)^2.[/MATH]
[MATH]\therefore \prod_{k=1}^6 (1 + x^k) = 4 *(1 + x)^2 * (1 + x^2)^2 =[/MATH]
[MATH]4 * \{(1 + x)(1 + x^2)\}^2 = 4(1 + x + x^2 + x^3)^2 = 4(2 + x + x^2)^2.[/MATH]
[MATH]\therefore x = \dfrac{-\ 1 - i\sqrt{3}}{2} \implies x^2 = \dfrac{-\ 1 + i\sqrt{3}}{2} \text { from above.}[/MATH]
[MATH]\therefore 4(2 + x + x^2)^2 = 4 * \left( 2 - \dfrac{-\ 1 - i\sqrt{3}}{2} + \dfrac{-\ 1 + i\sqrt{3}}{2} \right )^2= [/MATH]
[MATH]4 * \left (2 + \dfrac{-\ 1 - 1 - i\sqrt{3} + i \sqrt{3}}{2} \right)^2 = 4 * \left (2 + \dfrac{-\ 2}{2} \right )^2 = 4.[/MATH]
Which is what you got.

[MATH]\therefore x = \dfrac{-\ 1 + i\sqrt{3}}{2} \implies x^2 = \dfrac{-\ 1 - i\sqrt{3}}{2} \text { from above.}[/MATH]
[MATH]\therefore 4(2 + x + x^2)^2 = 4 * \left( 2 + \dfrac{-\ 1 + i\sqrt{3}}{2} + \dfrac{-\ 1 - i\sqrt{3}}{2} \right )^2= [/MATH]
[MATH]4 * \left (2 + \dfrac{-\ 1 - 1 + i\sqrt{3} - i \sqrt{3}}{2} \right)^2 = 4 * \left (2 + \dfrac{-\ 2}{2} \right )^2 = 4.[/MATH]
In other words, there are TWO answers, namely 4 and 64.

 

[Dr.Peterson]

What does it mean to combine the first two factors and then multiply them?

I understand that to have meant expanding [MATH](1+x)(1+x^2) = 1 + x + x^2 + x^3 = (1 + x + x^2) + (x^3) = 0 + 1 = 1[/MATH], which is the last step in my suggested method.

There is x a complex number.

@Vali, can you confirm my interpretation of your first line as meaning that x is either of the non-real solutions? The word "complex" does not mean that, but is commonly misused that way. How was the original exercise worded?

 

[Vali]

If alfa ... , then .... is:

 

[Dr.Peterson]

Very good. It explicitly says that alpha is a complex number that is not real. So the concerns about alpha being 1 are extraneous. Also, you hadn't mentioned that choices were given; best practice is to always include choices, because sometimes they affect the meaning of the problem, or how it can be solved. In this case, I don't think they do.