A Cubic Polynomial Transformed

$Let\quad f\left( x \right) ={ x }^{ 3 }+7{ x }^{ 2 }+2x+9=\left( x-\alpha \right) \left( x-\beta \right) \left( x-\gamma \right) \\ \\ \Rightarrow \cfrac { a }{ b } =\cfrac { \left( 3\alpha -2 \right) \left( 3\beta -2 \right) \left( 3\gamma -2 \right) }{ \left( 3\alpha +2 \right) \left( 3\beta +2 \right) \left( 3\gamma +2 \right) } \\ \\ =\cfrac { { \left( -3 \right) }^{ 3 }\left( \cfrac { 2 }{ 3 } -\alpha \right) \left( \cfrac { 2 }{ 3 } -\beta \right) \left( \cfrac { 2 }{ 3 } -\gamma \right) }{ { \left( -3 \right) }^{ 3 }\left( \cfrac { -2 }{ 3 } -\alpha \right) \left( \cfrac { -2 }{ 3 } -\beta \right) \left( \cfrac { -2 }{ 3 } -\gamma \right) } \\ \\ =\cfrac { f\left( \cfrac { 2 }{ 3 } \right) }{ f\left( \cfrac { -2 }{ 3 } \right) } \\ \\ =\cfrac { { \left( \cfrac { 2 }{ 3 } \right) }^{ 3 }+7{ \left( \cfrac { 2 }{ 3 } \right) }^{ 2 }+2\left( \cfrac { 2 }{ 3 } \right) +9 }{ { \left( \cfrac { -2 }{ 3 } \right) }^{ 3 }+7{ \left( \cfrac { -2 }{ 3 } \right) }^{ 2 }+2\left( \cfrac { -2 }{ 3 } \right) +9 } \\ \\ =\cfrac { 371 }{ 283 } \\ \\ \Rightarrow a+b=371+283=654$

Using Vieta's Formulas, we have:

• $\alpha+\beta+\gamma = -7$
• $\alpha\beta+\beta\gamma+\gamma\alpha = 2$
• $\alpha\beta\gamma = -9$

Now:

$\dfrac {(3\alpha - 2)(3\beta - 2)(3\gamma - 2)} {(3\alpha + 2)(3\beta + 2)(3\gamma + 2)}$

$= \dfrac {12(\alpha+\beta+\gamma)-18(\alpha\beta+\beta\gamma+\gamma\alpha)+27\alpha\beta\gamma-8}{12(\alpha+\beta+\gamma)+18(\alpha\beta+\beta\gamma+\gamma\alpha)+27\alpha\beta\gamma+8}$

$=\dfrac {12(-7)-18(2)+27(-9)-8}{12(-7)+18(2)+27(-9)+8} = \dfrac {371}{283}$

$\Rightarrow a = 371$ , $b = 283$ and $a+b = \boxed{654}$

The problem can also be solved by taking the title of the problem at face value:

Make the substitution $x\rightarrow \frac{x}{3}$ to get a new polynomial whose roots are $3\alpha ,3\beta ,3\gamma$ .

The given expression then reduces to $\frac{f(2)}{f(-2)}$ where $f(x)$ is the polynomial obtained after making the aforementioned substitution.

Of course, this is wholly equivalent to $Ayush$ $Verma's$ solution.

α+β+γ=-7 αβ+βγ+αγ=2 αβγ=-9 (3α-2)(3β-2)(3γ-2)=-371 (3α+2)(3β+2)(3γ+2)=-283 a/b= -7*53/-283 on comparing a=371 b=283 a+b=654

i did it by opening the brackets ad then solving using vietas formukla