Subtract : $3p^3-2q+r$ from $9p^2+2q-4r$ and add the result to $p^2-3p^3+p-r$.

Given :

The given expressions are $3p^3-2q+r$, $9p^2+2q-4r$ and $p^2-3p^3+p-r$.

To do :

We have to subtract  $3p^3-2q+r$ from $9p^2+2q-4r$ and add the result to $p^2-3p^3+p-r$.

Solution :

$(9p^2+2q-4r)-(3p^3-2q+r)+(p^2-3p^3+p-r)$

$= [9p^2+2q-4r-3p^3-(-2q)-r]+(p^2-3p^3+p-r)$

$=(-3p^3+9p^2+2q+2q-4r-r)+(p^2-3p^3+p-r)$

$=(-3p^3+9p^2+4q-5r)+(p^2-3p^3+p-r)$

$=-3p^3-3p^3+9p^2+p^2+4q-5r-r+p$

$=-6p^3+10p^2+p+4q-6r$

Therefore, the answer is $-6p^3+10p^2+p+4q-6r$.