Shall we use?

The roots of the polynomial : $2x^3-9x^2+11x-84$ are $a,b,c$ .

As we know from the General form of a polynomial equation, aka Vieta's Formula,

$a+b+c=-\frac{-9}{2}=\frac{9}{2}$

$ab+bc+ac=\frac{11}{2}$

$abc=-\frac{-84}{2}=\frac{84}{2}=42$

Therefore,

$X=a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ac)=\frac{81}{4}-11$

$Y=a(ab+bc+ac)+b(ab+bc+ac)+c(ab+bc+ac)=(a+b+c)(ab+bc+ac)=\frac{99}{4}$

$Z=abc(a+b+c)=189$

Thus, $X+Y+Z=\boxed{223}$

P.S. Here is a link to Wikipedia for Vieta's Formula. :)

Use Vieta's formulas:-
a+b+c=4.5, ......ab+bc+ac=5.5,......abc=42.
X=a^2+b^2+c^2=(a+b+c)^2 -2(ab+bc+ac)=4.5^2- 2*5.5=9.25..........9.25\\ Y=a^2(b+c)+b^2(c+a)+c^2(a+b)+3abc=a^2(4.5 - a)+b^2(4.5 - b)+c^2(4.5 - c)+3abc\\ \ \ \ =(a+b+c)(a^2+b^2+c^2) - (a+b+c)(a^2+b^2+c^2 - ab - bc - ac)\\ \ \ \ =(a+b+c)(ab+bc+ac)\ \ \ \=4.5*5.5=24.75...... =4.5*5.5=24.75......... 24.75\\ Z= abc(a+b+c)=42*4.5=189......................................................................189\\ X+Y+Z=9.25+24.75+189=223........................................................\color{#D61F06}{223} .

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

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

$Z=abc(a+b+c)$

Then use Vieta's formulas to solve for each.

The problem can easily be solved by Vieta's Formulas

Kindly refer to the link given if you are not familiar with these formulae.

By Vieta's, We have, In a cubic Polynomial

$P(x)={ax}^{3}+{bx}^{2}+cx+d$

Here, a=2,b=-9,c=11,d=-84

a+b+c = $\dfrac{-b}{a}$

ab+bc+ca= $\dfrac{c}{a}$

abc= $\dfrac{-d}{a}$

1) ${a}^{2}+{b}^{2}+{c}^{2}$ = ${(a+b+c)}^{2}-2(ab+bc+ca)$

= ${\dfrac{-b}{a}}^{2}-2\dfrac{c}{a}$

= ${\dfrac{9}{2}}^{2}-2\dfrac{11}{2}$

$\dfrac{81}{4}$ - 11

= $\dfrac{37}{4}$

Now, I encourage you to Try the other parts on your own. I will give Hints.

2) ${a}^{2}b+abc+{a}^{2}c+{b}^{2}a+{b}^{2}c+abc+abc+{c}^{2}a+b{c}^{2}$ Can be factorized as (a+b+c)(ab+bc+ca). The answer comes out to be $\dfrac{99}{4}$

3) ${a}^{2}bc+{b}^{2}ca+{c}^{2}ab$ can be factorized as (a+b+c)(abc) The answer comes out to be 189.

Adding these, We get 223.