Clever

Since $p$ , $q$ and $r$ are the roots of $x^3+x^2+x-8=0$ , then $p^3+p^2+p=8$ , $q^3+q^2+q=8$ and $r^3+r^2+r=8$ .

Therefore,

$p^3+2p^2+p+q^3+2q^2+r^3+2r^2+r$

$=p^3+p^2+p+q^3+q^2+r^3+r^2+r+p^2+q^2+r^2$

$=8+8+8+p^2+q^2+r^2=24+p^2+q^2+r^2$

We now need to find $=p^2+q^2+r^2$ .

Since $p$ , $q$ and $r$ are the roots of $x^3+x^2+x-8=0$ , then:

$x^3+x^2+x-8=(x-p)(x-q)(x-r)$

$\quad \quad\ \quad \quad \quad \quad = x^3 - (p+q+r)x^2 + (pq+qr+rp)x - pqr$

$\Rightarrow p+q+r = -1$

$\quad pq+qr+rp = 1$

Now we have:

$(p+q+r)^2 = p^2+q^2+r^2 + 2(pq+qr+rp)$

$(-1)^2 = p^2+q^2+r^2 + 2(1)$

$\Rightarrow p^2+q^2+r^2 = 1 -2 = -1$

Therefore, $p^3+2p^2+p+q^3+2q^2+r^3+2r^2+r=24-1=\boxed{23}$

I think the problem is easier knowing Newton´s Sums.

Lets see ${ p }^{ 3 }+{ 2p }^{ 2 }+{ p }+{ q }^{ 3 }+{ 2q }^{ 2 }+q+{ r }^{ 3 }+{ 2r }^{ 2 }+{ r }$ can be written as $(p+q+r)+2({ p }^{ 2 }+{ q }^{ 2 }+{ r }^{ 2 })+({ p }^{ 3 }+{ q }^{ 3 }+{ r }^{ 3 })$ .

Say ${ P }_{ 1 }=p+q+r,{ P }_{ 2 }={ p }^{ 2 }+{ q }^{ 2 }+{ r }^{ 2 },{ P }_{ 3 }={ p }^{ 3 }+{ q }^{ 3 }+{ r }^{ 3 }$ .Now according to Newton´s Sums, we know:

${ P }_{ 1 }+1=0\Rightarrow { P }_{ 1 }=-1$

${ P }_{ 2 }+{ P }_{ 1 }+2=0\Rightarrow { P }_{ 2 }=-1$

${ P }_{ 3 }+{ P }_{ 2 }+{ P }_{ 1 }-24=0\Rightarrow { P }_{ 3 }=26$ .

So $(p+q+r)+2({ p }^{ 2 }+{ q }^{ 2 }+{ r }^{ 2 })+({ p }^{ 3 }+{ q }^{ 3 }+{ r }^{ 3 })=(-1)+2(-1)+26=\boxed { 23 }$ .

Since, x^3+x^2+x-8=0 is polynomial equation with leading coefficient 1, therefore, sum of the roots=p+q+r= -(coefficient of x^2/coefficient of x^3)=-(1)/(1)=- 1 and product of roots taken two at a time=pq+qr+rp=(coefficient of x/coefficient of x^3)= (1)/(1)=1 My remaining steps are the same as that of Chew-Seong Cheong

Since p,q,r are the roots of the given equation therefore we can write x=p,x=q & x=r...from the given equation we get p+q+r=-1 & pq+qr+pr=1...now the expression whose value is to be found out can be written as x^3 + 2 x^2 + x + x^3 + 2 x^2 + x + x^3 + 2 x^2 +x = 3 x^3 + 6 x^2 + 3x....we have x^3 + x^2 + x=8....So divide (3 x^3 + 6 x^2 + 3x) by (x^3 + x^2 + x) using long division....we get quotient as 3 and remainder as 3 x^2....so (3 x^3 + 6 x^2 +3x)=((x^3 + x^2 + x) * 3 ) + 3 x^2...= 8*3 + 3 x^2 = 24 + x^2 + x^2 + x^2 =24 + p^2 +q^2 + r^2 = 24+ ((p+q+r)^2 -2(pq+qr+pr)) = 24+1-2 = 23...