Oh no not cubics!

Hey bro, I'm afraid you are wrong with the value of $r$ . It should be: $r=\dfrac{a^2+b^2+c^2}{abc}=\dfrac{(a+b+c)^2-2(ab+ac+bc)}{abc}=\dfrac{(2)^2-2(3)}{4}=-\dfrac{1}{2}$ . You forgot to multiply by $2$ . Hence, the correct answer should be $\dfrac{-\dfrac{1}{2}}{-\dfrac{7}{16}}=\dfrac{8}{7}$ , and $8+7=\boxed{15}$ . Check it please, nice problem.

After reviewing this, I agree that the expression should be $- \frac{8}{7}$ .

Those who previously answered -1, 1 or 15 have been marked correct. I've also rephrased the question so that the answer is now 15.

The answer should be 1. I'll prove my claim.

${ 7x }^{ 3 }\quad -\quad 14{ x }^{ 2 }\quad +\quad 21x\quad -\quad 28\quad has\quad roots\quad a,\quad b\quad and\quad c.\\ Given\quad that\quad { x }^{ 3 }\quad +\quad { rx }^{ 2 }\quad +\quad sx\quad +\quad t\quad has\quad roots\quad \frac { a }{ bc } ,\quad \frac { b }{ ac } \quad and\quad \frac { c }{ ab } .\\ We\quad know\quad that\quad a\quad +\quad b\quad +\quad c\quad =\quad 2,\quad ab\quad +\quad bc\quad +\quad ac\quad =\quad 3,\quad and\quad abc\quad =\quad 4.\\ \\ Let\quad f(x)\quad =\quad { x }^{ 3 }\quad +\quad { rx }^{ 2 }\quad +\quad sx\quad +\quad t\\ \\ Sum\quad of\quad roots\quad of\quad f(x)\quad =\quad \frac { -r }{ 1 } \\ \frac { a }{ bc } +\quad \frac { b }{ ac } +\quad \frac { c }{ ab } \quad =\quad -r\\ \frac { { a }^{ 2 }\quad +\quad { b }^{ 2 }\quad +\quad { c }^{ 2 } }{ abc } =\quad -r\\ \frac { { (a\quad +\quad b\quad +\quad c) }^{ 2 }\quad -\quad 2(ab\quad +\quad bc\quad +\quad ac) }{ abc } =\quad -r\\ Substituting\quad the\quad values:\quad -\\ \frac { { 2 }^{ 2 }\quad -\quad 2(3) }{ 4 } =\quad -r\\ \therefore \quad -r\quad =\quad \frac { -1 }{ 2 } \quad \Longrightarrow \quad r\quad =\quad \frac { 1 }{ 2 } \\ Sum\quad of\quad product\quad of\quad roots\quad of\quad f(x)\quad =\quad \frac { s }{ 1 } \\ \frac { a }{ bc } \times \frac { b }{ ac } \quad +\quad \frac { b }{ ac } \times \frac { c }{ ab } \quad +\quad \frac { c }{ ab } \times \frac { a }{ bc } \quad =\quad s\\ \frac { 1 }{ { a }^{ 2 } } \quad +\quad \frac { 1 }{ { b }^{ 2 } } \quad +\quad \frac { 1 }{ { c }^{ 2 } } =\quad s\\ \frac { { (ab) }^{ 2 }\quad +\quad { (bc) }^{ 2 }\quad +\quad { (ac) }^{ 2 }\quad }{ { (abc) }^{ 2 } } =\quad s\\ \frac { { (ab\quad +\quad bc\quad +\quad ac) }^{ 2 }\quad -\quad 2abc(a\quad +\quad b\quad +\quad c) }{ { (abc) }^{ 2 } } =\quad s\\ Substituting\quad the\quad values:\quad -\\ \frac { { 3 }^{ 2 }\quad -\quad (2\quad \times \quad 4\quad \times \quad 2) }{ 16 } =\quad s\quad \Longrightarrow \quad s\quad =\quad \frac { -7 }{ 16 } \\ \therefore \quad \frac { r }{ s } \quad =\quad \frac { \frac { 1 }{ 2 } }{ \frac { -7 }{ 16 } } \quad =\quad \frac { 1 }{ 2 } \times \quad \frac { 16 }{ -7 } \quad =\quad \frac { 8 }{ -7 } \\ \therefore \quad \frac { p }{ q } \quad =\quad \frac { 8 }{ -7 } \quad \Longrightarrow \quad p\quad +\quad q\quad =\quad \boxed { 1 } \\ \\ \\$

Yes its wrong