Francisco's Cubic Roots

Note that $P(x)=(x-r_1)(x-r_2)(x-r_3)$ and $r_k^2+r_k+1=(w-r_k)(w^2-r_k)$ , where $w$ is a primitive cubic root of unity.

Then, our product is $(w-r_1)(w^2-r_1)(w-r_2)(w^2-r_2)(w-r_3)(w^2-r_3)$ , or simply $P(w)P(w^2)$ .

$P(w)P(w^2)=(w^3+3w+1)(w^6+3w^2+1)\\ P(w)P(w^2)=(2+3w)(2+3w^2)\\ P(w)P(w^2)=4-2(3)+3^2\\ P(w)P(w^2)=\boxed {7}$

$\begin{aligned} \prod_{k=1}^3 \left(r_k^2+r_k+1\right) & = \prod_{k=1}^3 \left(\frac{\color{#3D99F6} {r_k^3} -1}{r_k-1}\right) \\ & = \prod_{k=1}^3 \left(\frac{\color{#3D99F6} {(-3r_k-1)} -1}{r_k-1}\right) \quad \quad \color{#3D99F6} {[r_k^3 + 3r_k + 1 = 0]} \\ & = - \prod_{k=1}^3 \left(\frac{3r_k + 2}{r_k-1}\right) \\ & = - \frac{(3r_1 + 2)(3r_2 + 2)(3r_3 + 2)}{(r_1-1)(r_2-1)(r_3-1)} \\ & = - \frac {27r_1r_2r_3 + 18(r_1r_2+r_2r_3+r_3r_1)+12(r_1+r_2+r_3)+8} {r_1r_2r_3-(r_1r_2+r_2r_3+r_3r_1) + r_1+r_2+r_3 - 1} \\ & = - \frac {27(\color{#3D99F6}{-1})+18(\color{#3D99F6}{3}) +12(\color{#3D99F6}{0})+8} {\color{#3D99F6}{-1}+\color{#3D99F6}{3} +\color{#3D99F6}{0}-1} \quad \quad \color{#3D99F6} {\text{[Vieta's formulas]}} \\ & = \frac{35}{5} = \boxed{7} \end{aligned}$

$x^3+3x+1 = (x^2 + x + 1)(x-1)+(3x+2) \implies r^2 + r + 1 = \frac{3r+2}{1-r}$
$\prod{(r^2+r+1)} = \frac{\prod{(3r+2)}}{\prod{(1-r)}} = \frac{-27 f(-2/3)}{f(1)} = \boxed{7}$

I actually managed to get the answer by expanding the product itself. Sometimes, bulldozing through is an option too. It took more than half an hour though.

For easy typing, let $r_1 = a, r_2 = b$ and $r_3 = c$

Therefore:

$\displaystyle \prod_{k=1}^3(r_k^2+r_k+1) = (a^2 + a + 1)(b^2 + b + 1)(c^2 + c + 1)$

From Vieta's formulas (go check it out if you do not know this),

$a + b + c = 0$

$ab + ac + bc = 3$

$abc = -1$

Expand the entire product and you'll get an expression with 27 terms:

$(a^2 + a + 1)(b^2 + b + 1)(c^2 + c + 1)$

$= a^2b^2c^2 + a^2bc^2 + a^2c^2 + ab^2c^2 + abc^2 + ac^2 + b^2c^2 + bc^2 + c^2$

$+ a^2b^2c +a^2bc + a^2c + ab^2c + abc + ac + b^2c + bc + c$

$+ a^2b^2 + a^2b + a^2 + ab^2 + ab + a + b^2 + b + 1$

Sort out the entire mess and split it as below:

$(a^2 + a + 1)(b^2 + b + 1)(c^2 + c + 1)$

$= a^2b^2c^2 + abc + ab + ac + bc + a + b + c + 1$

$+ a^2b^2c + a^2bc^2 + ab^2c^2 + a^2bc + ab^2c + abc^2$

$+ a^2 + b^2 + c^2$

$+ a^2b^2 + a^2c^2 + b^2c^2$

$+ a^2b + a^2c + ab^2 + b^2c + ac^2 + bc^2$

Now, I'll solve the split rows one by one.

The first row, you can substitute the values directly from the equations above:

$a^2b^2c^2 + abc + ab + ac + bc + a + b + c + 1$

$= (abc)^2 + abc + (ab + ac + bc) + (a + b + c) + 1$

$= 1 - 1 + 3 + 0 + 1 = \boxed{4}$

For the second row, you can factorize $abc$ :

$a^2b^2c + a^2bc^2 + ab^2c^2 + a^2bc + ab^2c + abc^2$

$= abc (ab + ac + bc + a + b + c)$

$= -1 (3 + 0) = \boxed{-3}$

For the third row, square the first equation above:

$(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc$

Therefore,

$a^2 + b^2 + c^2 = (a + b + c)^2 - 2 (ab + ac + bc)$

$= 0^2 -2 (3) = \boxed{-6}$

For the fourth row, square the second equation above:

$(ab + ac + bc)^2 = a^2b^2 + a^2c^2 + b^2c^2 + 2a^2bc + 2ab^2c + 2abc^2$

Therefore,

$a^2b^2 + a^2c^2 + b^2c^2$

$= (ab + ac + bc)^2 - 2a^2bc - 2ab^2c - 2abc^2$

$= (ab + ac + bc)^2 - 2abc (a + b + c)$

$= 3^2 - 2 (-1) (0) = \boxed{9}$

The final row is a bit tricky:

$a^2b + a^2c + ab^2 + b^2c + ac^2 + bc^2$

$= ab (a + b) + ac (a + c) + bc (b + c)$

From equation 1,

$a + b + c = 0 \implies a + b = -c, a + c = -b, b + c = -a$

Substitute the three equations here, and you get:

$ab (-c) + ac (-b) + bc (-a)$

$= -abc - abc - abc$

$= -3 abc = -3(-1) = \boxed{3}$

With the values of all 5 rows obtained, we can get the product:

$(a^2 + a + 1)(b^2 + b + 1)(c^2 + c + 1) = 4 - 3 - 6 + 9 + 3$

$\displaystyle \prod_{k=1}^3(r_k^2+r_k+1) = \boxed{7}$

Note: I used this method because I had no other better way to do it, and I didn't want to view solutions. There are much better solutions listed above. I wrote my solution only because I didn't want my efforts to go to waste

Write $P(x) = x^3 + 3 x + 1 = (x-1)(x^2 + x +1) + (2+3x)$ and since $r_k$ are the roots of $P(x)$ , then for each $k$ $r_{k}^2 + r_{k} +1 = \frac{ 2+3r_{k}}{1-r_{k}}$

Hence, we have $\prod_{k=1}^{3} (r_{k}^2 + r_{k} +1 )= \frac{8 + 12(r_1 + r_2 + r_3 ) + 18 (r_1 r_2 + r_2 r_3 + r_1 r_3 ) + 27 r_1 r_2 r_3 }{1-(r_1 + r_2 + r_3) + (r_1 r_2 + r_2 r_3 + r_1 r_3 ) - r_1 r_2 r_3}$ From $P(x)$ , we have $r_1 + r_2 + r_3 = 0$ , $r_1 r_2 + r_2 r_3 + r_1 r_3 =3$ , and $r_1 r_2 r_3 = -1$ . Then $\prod_{k=1}^{3} (r_{k}^2 + r_{k} +1) = \frac{35}{5} = 7$

Since $(r_k^2 + r_k + 1)(r_k - 1) = r_k^3 - 1 = -3r_k - 2$ , the asked product is equal with the following one: $\prod_{k=1}^{3}\frac{-3r_k - 1}{r_k - 1}$ . Now, let the $y = x - 1$ a subtitution that transform the original ecuation into $y^3 + 3y^2 + 6y + 5 = 0$ and, according to Viete formulas, we have $y_1y_2y_3 = -5$ , thus $(r_1 - 1)(r_2 - 1)(r_3 - 1) = -5$ . Proceeding in the same manner with the subtitution $t = -3x - 2$ , we get the equation $t^3 + 6t^2 + 39t + 35 = 0$ , so that $t_1t_2t_3 = -35$ . Now, the asked product is $\frac{-35}{-5} = 7$ .

$\prod_{i=1}^{3} (x - r_i) = x^3 + 3x + 1 = P(x)$ .

$\prod_{i=1}^{3} (r^{2}_i + r_i + 1) = \frac {\prod_{i=1}^{3} (1 - r^3_i)}{\prod_{i=1}^{3}(1 - r_i)} = \frac {Q(1)}{P(1)}$ , where, $Q(x) = \prod_{i=1}^{3}(x - r^{3}_i)$ .

From P(x) we get, $r_1 + r_2 + r_3 = 0, r_1r_2 + r_2r_3 + r_3r_1 = 3$ and $r_1r_2r_3 = -1$ . Using these three equations, following are easily calculated:

$r^{3}_1 + r^{3}_2 + r^{3}_3 = -3, r^{3}_1r^{3}_2 + r^{3}_2r^{3}_3 + r^{3}_3r^{3}_1 = 30$ and $r^{3}_1r^{3}_2r^{3}_3 = -1$ .

Therefore, $Q(x) = x^3 + 3x^2 + 30x + 1$ .

$\frac {Q(1)}{P(1)} = \frac {35}{5} = 7$