<s>[INST] How find the $(x^2_{1}+x_{1}x_{2}+x^2_{2})(x^2_{2}+x_{2}x_{3}+x^3_{3})(x^2_{3}+x_{3}x_{1}+x^2_{1})$ Let  $x_1$, $x_2$, and $x_3$ be the roots of the equation
$$4x^3-6x^2+7x-8=0.$$
Find this value:
$$(x_1^2+x_1x_2+x_2^2)(x_2^2+x_2x_3+x_3^3)(x_3^2+x_3x_1+x_1^2)$$
My try:
$$x_1+x_2+x_3=\frac 3 2$$
$$x_1 x_2+x_2 x_3+x_1 x_3=\frac 7 4$$
$$x_1 x_2 x_3=2$$
Now I have
$$x_1^2+x_1x_2+x_2^2=(x_1+x_2)^2-x_1x_2=(\frac 3 2-x_3)^2-\frac 2 {x_3}.$$
But this is very ugly, and I think this problem should have a cleaner solution. Thanks.
 [/INST] Assuming that we need to calculate $$(x^2_1+x_1x_2+x^2_2)(x^2_2+x_2x_3+x^2_3)(x^2_3+x_3x_1+x^2_1)$$
Let $\displaystyle x^2_1+x_1x_2+x^2_2=y_3,x^2_2+x_2x_3+x^2_3=y_1, x^2_3+x_3x_1+x^2_1=y_2$
So, $\displaystyle y_3=x^2_1+x_1x_2+x^2_2=(x_1+x_2)^2-x_1x_2$
$=\left(\frac32-x_3\right)^2-\frac2{x_3}$ as $x_1+x_2+x_3=\frac32$ and  $x_1 x_2 x_3=2$
On re-arrangement we have,  $\displaystyle 4x_3^3-12x_3^2+x_3(9-4y_3)-8=0\ \ \ \  (1)$
Again as $x_3$ is a root of the given equation, $4x_3^3-6x_3^2+7x_3-8=0\ \ \ \ (2)$
$(2)-(1)\implies 6x_3^2+x_3\{7-(9-4y_3)\}=0\iff 2x_3(3x_3+2y_3-1)=0$
As $0$ does not satisfy the given equation $x_3\ne0$
$\displaystyle \implies 3x_3+2y_3-1=0\iff x_3=\frac{1-2y_3}3$
Putting the value of $x_3$ in $(2)$ we get,
$$4\left(\frac{1-2y_3}3\right)^3-6\left(\frac{1-2y_3}3\right)^2+7\left(\frac{1-2y_3}3\right)-8=0$$
On re-arrangement we have,  
$\displaystyle  4(2y_3)^3+(\cdots)y_3^2+(\cdots)y_3+3^3\cdot8-3^3\cdot7+3\cdot6-4=0$
$\displaystyle \implies 32y_3^3+(\cdots)y_3^2+(\cdots)y_3+167=0\ \ \ \ (3)$
As the values of $y_1,y_2,y_3$ are symmetric, we shall reach at the same equation $(3)$ if we start with $y_1$ or $y_2$
$\displaystyle \implies y_1,y_2,y_3$ are the roots equation $(3)$
Using Vieta's Formula, $\displaystyle y_1y_2y_3=-\frac{167}{32}$ 
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