# Problem involving cube roots of unity

Given that $$\frac{1}{a+\omega}+\frac{1}{b+\omega}+\frac{1}{c+\omega}=2\omega^2\;\;\;\;\;(1)$$ $$\frac{1}{a+\omega^2}+\frac{1}{b+\omega^2}+\frac{1}{c+\omega^2}=2\omega\;\;\;\;\;(2)$$

Find $$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=?\;\;\;\;\;\;(3)$$

I tried adding the given two equations, and simplified them. I'll show the working for one term here. $$\frac{1}{a+\omega}+\frac{1}{a+\omega^2}=\frac{2a-1}{a^2-a+1}=\frac{(2a-1)(a+1)}{(a^3+1)}=\frac{2a^2+a-1}{a^3+1}$$ $$\frac{1}{a+1}=\frac{a^2-a+1}{a^3+1}=\frac{1}{2}\left[\frac{(2a^2+a-1)-3(a-1)}{a^3+1}\right]=\frac12\left[\frac{2a^2+a-1}{a^3+1}\right]-\frac32\left[\frac{a-1}{a^3+1}\right]$$

Now, I cannot eliminate that extra term which I get at the end of the above expression. Is there hope beyond this? Or is there a better alternative?

• I suppose (using the title) that $\omega^3=1$, no? Nov 20 '14 at 13:10
• Yes.. $\omega^3=1$, $1+\omega+\omega^2=0$ Nov 20 '14 at 13:11

Clearly, the two of the three roots of $$\frac1{a+x}+\frac1{b+x}+\frac1{c+x}=\frac2x$$ are $\omega,\omega^2$

On simplification, $$2x^3+2(a+b+c)x^2+2(ab+bc+ca)x+2abc=3x^3+2x^2(a+b+c)+x(ab+bc+ca)$$

$$\iff x^3+x^2(0)-(ab+bc+ca)x-2abc=0$$

If $a$ is the third root, using Veita's formula $$a+\omega^2+\omega=-\dfrac01$$

Hope the rest should be easy to deal with.

• @Tejas, So, we have $$\sum\frac1{a+1}=\frac21$$ Nov 21 '14 at 8:07

Put $A(x)=(x+a)(x+b)(x+c)=x^3+ux^2+vx+w$. Then $$\frac{A^{\prime}(x)}{A(x)}=\frac{1}{x+a}+\frac{1}{x+b}+\frac{1}{x+c}$$ Hence we get $A^{\prime}(\omega)=2\omega^2 A(\omega)$ and $A^{\prime}(\omega^2)-2\omega A(\omega^2)=0$, or if $Q(x)=xA^{\prime}(x)-2A(x)$, that $Q(\omega)=Q(\omega^2)=0$. Hence $Q(x)$ is divisible by $(x-\omega)(x-\omega^2)=x^2+x+1$. We have $$Q(x)=x^3-vx-2w=(x-1)(x^2+x+1)+(-vx+1-2w)$$ Hence $v=0$, $w=1/2$, and $Q(x)=(x-1)(x^2+x+1)$, $A'(1)=2A(1)$, and we are done.

I would like to add to the above answer, the symmetry put a spark in my brain and i came up with this.

As in answer stated by peiter.

Assume

$X=\large\frac{1}{a+1}+\large\frac{1}{b+1}+\large\frac{1}{c+1}$ $\space\space\space\space\space\space\space\space$ and

$\large\frac{1}{a+\omega}+\large\frac{1}{b+\omega}+\large\frac{1}{c+\omega}=2\omega^2$

$\large\frac{1}{a+\omega^2}+\large\frac{1}{b+\omega^2}+\large\frac{1}{c+\omega^2}=2\omega$

Adding the three equation and considering first term related to "$a$"

$X+2\omega+2\omega^2=\large\frac{1}{a+1}+\large\frac{1}{a+\omega}+\large\frac{1}{a+\omega^2}+other\space terms$

simplifies to,

$X+2\omega+2\omega^2=\large\frac{3a^2}{a^3+1}+\large\frac{3b^2}{b^3+1}+\large\frac{3c^2}{c^3+1}$

Now as according to solution given by pieter $X=2$

Therefore

$\large\frac{3a^2}{a^3+1}+\large\frac{3b^2}{b^3+1}+\large\frac{3c^2}{c^3+1}$ must equal zero.

If somehow you can prove that , the answer may become simpler.(i.e., to prove $X=2$)

I would suggest trying to multiplying and dividing the express by $\omega$ and $\omega^2$ to arrive at other expressions that may yield this result.

My original answer did not use the nice symmetry. Kelenner gave a better answer. Based on that, I have made yet another answer, using essentially the same method as Kelenner did, but with a different definition of the same function which is (for me) easier to understand. When I first wrote this answer, I did not realize I was using exactly the same function, otherwise I would not have written this down, but now that it is here I'll leave it, because it might be easier to read for some people.

Define the function $Q$ by $$Q(x)=\left(\left(\frac{1}{a+x}+\frac{1}{b+x}+\frac{1}{c+x}\right) x-2\right)(a+x)(b+x)(c+x)$$ From the given equations: $$Q(\omega)=0,\quad Q(\omega^2)=0.$$ The function $Q$ can be much simplified (with simple manipulations): $$Q(x)=x^3-2abc$$ Because $\omega$ and $\omega^2$ are zeroes of $Q$, $Q$ must have a factor $(x^2+x+1)$. Following symbolic division, this can only be true if $(1-2abc)=0$, and then $$Q(x)=(x^2+x+1)*(x-1)$$ So, $Q(1)=0$, and from the original definition of $Q$ it must follow that $$\frac{1}{a+x}+\frac{1}{b+x}+\frac{1}{c+x}=2$$.

Multiply all equations by the product of the denominators ($(a-\omega)(b-\omega)(c-\omega)$ in (1), $(a-\omega^2)(b-\omega^2)(c-\omega^2)$ in (2) etc.), to get rid of the fractions. The first two equations then simplify to (remember $\omega^3=1$):

$$0=(ab+ac+bc)+(2abc-1)\omega^2$$ $$0=(ab+ac+bc)+(2abc-1)\omega$$ Because $\omega\neq\omega^2$, these can only both be true if $2abc-1=0$. In that case both equations become $$ab+ac+bc=0.$$ If I replace the questionmark in your equation (3) by $X$, the equation becomes (after multiplying with the three terms): $$X=\frac{(b+1)(c+1)+(a+1)(c+1)+(a+1)(b+1)}{(a+1)(b+1)(c+1)}$$ Expanded: $$X=\frac{ab+bc+ac+2a+2b+2c+3}{abc+ab+ac+bc+a+b+c+1}$$ Use the fact that $ab+ac+bc=0$ and $2abc-1=2$: $$X=\frac{2a+2b+2c+3}{a+b+c+\frac{3}{2}}=2$$ There must be a nicer way to show this, considering the nice symmetry in the problem, but I can't see that now.