Clever Manipulations 1

By Vieta, we have $p+q+r=2$ and $pq+qr+pr=2$ so $p^2+q^2+r^2=(p+q+r)^2-2(pq+pr+qr)=0$ . So we can substitute $p^2=-q^2-r^2$ into $p^2-2qr$ to get that

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

We also have that $q+r=2-p$ so $p^2-2qr=-(2-r)^2$ . Doing similarly for the other two, we get the desired product as

$-[(2-r)(2-p)(2-q)]^2$

Rather than expanding this, we note that $x^3-2x^2+2x-11=(x-p)(x-q)(x-r)$ since $p,q$ , and $r$ are the three roots. Thus substituting $x=2$ matches the expression above so

$2^3-2(2^2)+2(2)-11=-7=(2-r)(2-p)(2-q)$

so $-[(2-r)(2-p)(2-q)]^2=\boxed{-49}$ .

Since $pqr=11$ $\Rightarrow (p^2-2qr)=(p^2-\frac{22}{p})$ So our required expression becomes- $\frac{(p^3-22)(q^3-22)(r^3-22)}{pqr}$ $=\frac{(p^3-22)(q^3-22)(r^3-22)}{11}=\frac{P}{11}(say)$ Now we need to find $P$ . One way we can do that is to put $x=(y+22)^{\frac{1}{3}}$ so that the new cubic obtained has $P$ as the product of its three roots- $(p^3-22),(q^3-22)$ & $(r^3-22)$ . I will directly write the new equation which is: $y^3+37y^2+415y+539=0$ Now from here $P=-539$ $\huge \Rightarrow \frac{P}{11}=\boxed{-49}$

$\left( { p }^{ 2 }-2qr \right) \left( { q }^{ 2 }-2pr \right) \left( { r }^{ 2 }-2pq \right)$

Which give us

$-7{ \left( pqr \right) }^{ 2 }-2\left( { p }^{ 3 }{ q }^{ 3 }+{ q }^{ 3 }{ r }^{ 3 }+{ r }^{ 3 }{ p }^{ 3 } \right) +4pqr\left( { p }^{ 3 }+{ q }^{ 3 }+{ r }^{ 3 } \right) ......\left( i \right)$

Now By Vieta we have

$pqr=11\\ p+q+r=2\\ pq+qr+pr=2$

We know

$\left( a+b+c \right) ^{ 3 }={ a }^{ 3 }+{ b }^{ 3 }+{ c }^{ 3 }+3\left[ \left( a+b+c \right) \left( ab+bc+ca \right) -abc \right]$

Putting $a=p\quad ,\quad b=q\quad and\quad c=r$

We get $p^{ 3 }+{ q }^{ 3 }+{ r }^{ 3 }=29$

Similarly by putting

$a=pq\quad b=qr\quad and\quad c=pr$

we get $pq^{ 3 }+{ qr }^{ 3 }+{ rp }^{ 3 }=239$

now plugging all these values in $\quad \quad ....(i)$

$we\quad get\quad -7{ \left( 11 \right) }^{ 2 }-2\left( 239 \right) +44\left( 29 \right)$

$\huge{\Rightarrow -\boxed{49}}$

I was making mistakes using the other methods for some reason so I settled for this alternative solution using Newton's sums.

Same solution here too!

I see many have solved it but due to the enormous length of the solution, few wanna post the solution.

Thst applies for me too.