It is a factor if $$a{x^3} + bx + c = \left( {{x^2} + px + 1} \right)\left( {ax + c} \right)$$
$$\eqalign{ & = a{x^3} + \left( {c + ap} \right){x^2} + \left( {pc + a} \right)x + c \cr & \Rightarrow \,\,c + ap = 0,pc + a = b \cr & \Rightarrow \,\,p = \frac{{ - c}}{a} = \frac{{b - a}}{c}. \cr} $$

$$\eqalign{ & \alpha - 1 > 0,\beta - 1 > 0 \cr & \Rightarrow \,\,\alpha + \beta - 2 > 0\,\,{\text{and }}\alpha \beta - \left( {\alpha + \beta } \right) + 1 > 0. \cr & \alpha - 2 < 0,\beta - 2 > 0 \cr & \Rightarrow \,\,\alpha \beta - 2\left( {\alpha + \beta } \right) + 4 < 0. \cr & \alpha - 3 < 0,\beta - 3 < 0 \cr & \Rightarrow \,\,\alpha + \beta - 6 < 0\,\,{\text{and }}\alpha \beta - 3\left( {\alpha + \beta } \right) + 9 > 0. \cr} $$
Also $$D > 0.$$  Thus obtain six inequations and solve them together.

Given, $$\alpha < \beta ,c < 0,b < 0,$$
$$\therefore \alpha + \beta = - b < 0\,\,{\text{and }}\alpha \beta = c < 0$$
Clearly, $$\alpha$$ and $$\beta$$ have opposite signs and $$\alpha < \beta$$
$$\eqalign{ & \therefore \alpha < 0\,\,{\text{and }}\beta > 0 \cr & \Rightarrow \alpha < 0 < \beta \cr} $$
Further, $$\alpha + \beta < 0$$
$$\eqalign{ & \Rightarrow \beta < - \alpha \cr & \Rightarrow \left| \beta \right| < \left| { - \alpha } \right| \cr & \Rightarrow \beta < \left| \alpha \right|\left( {\beta > 0 \Rightarrow \left| \beta \right| = \beta } \right) \cr} $$
Hence, $$\alpha < 0 < \beta < \left| \alpha \right|$$

We have
$$\eqalign{ & \frac{{\left( {{a_1} + {a_2} + ..... + {a_{n - 1}} + 2{a_n}{\mkern 1mu} } \right)}}{n} \geqslant {\left( {{a_1}{a_2}.....{a_{n - 1}}2{a_n}} \right)^{\frac{1}{n}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\text{Using A}}{\text{.M}}{\text{.}} \geqslant \,{\text{G}}{\text{.M}}{\text{.}}} \right] \cr & \Rightarrow \,\,{a_1} + {a_2} + {a_3} + ..... + {a_{n - 1}} + 2{a_n}{\mkern 1mu} \geqslant n{\left( {2c} \right)^{\frac{1}{n}}} \cr} $$

$$f\left( {x,y} \right) = \frac{1}{2}\left[ {{{\left( {x - 2y} \right)}^2} + {{\left( {2y - 3z} \right)}^2} + {{\left( {3z - x} \right)}^2}} \right] \geqslant 0.$$

Here, $$m$$ and $$n$$ are the roots of equation.
$$\eqalign{ & \left( {x + p} \right)\left( {x + q} \right) - k = 0 \cr & {x^2} + x\left( {p + q} \right) + pq - k = 0\,\,\,\,.....\left( {\text{i}} \right) \cr} $$
If $$m$$ and $$n$$ are the roots of equation, then
$$\eqalign{ & \left( {x - m} \right)\left( {x - n} \right) = 0 \cr & \therefore {x^2} - \left( {m + n} \right)x + mn = 0\,\,\,.....\left( {{\text{ii}}} \right) \cr} $$
Now equation (i) should be equal to equation (ii),
$$\left( {m + n} \right) = - \left( {p + q} \right)\,\,{\text{and }}mn = pq - k$$
Now, we have to find roots of $$\left( {x - m} \right)\left( {x - n} \right) + k = 0$$
$$\eqalign{ & {x^2} - \left( {m + n} \right)x + mn + k = 0 \cr & {x^2} + \left( {p + q} \right)x + \left( {pq - k} \right) + k = 0 \cr & {x^2} + \left( {p + q} \right)x + pq = 0 \cr & {x^2} + px + qx + pq = 0 \cr & x\left( {x + p} \right) + q\left( {x + p} \right) = 0 \cr & \therefore x + q = 0\,\,{\text{or }}x + p = 0 \cr & \therefore x = - q\,\,{\text{and }}x = - p \cr} $$

The given equation is
$$\eqalign{ & {m^2}{x^2} + \left( {2m - {m^2}} \right)x + 3 = 0 \cr & \therefore \alpha + \beta = - \frac{{2m - {m^2}}}{{{m^2}}} = \frac{{m - 2}}{m}\,{\text{and }}\alpha \beta = \frac{3}{{{m^2}}} \cr & {\text{Now,}} \cr & \frac{\alpha }{\beta } + \frac{\beta }{\alpha } = \frac{4}{3} \cr & \Rightarrow \frac{{{\alpha ^2} + {\beta ^2}}}{{\alpha \beta }} = \frac{4}{3} \cr & \Rightarrow \frac{{{{\left( {\alpha + \beta } \right)}^2} - 2\alpha \beta }}{{\alpha \beta }} = \frac{4}{3} \cr} $$
Substituting the values, we get
$$\eqalign{ & \frac{{{{\left( {\frac{{m - 2}}{m}} \right)}^2} - 2 \cdot \frac{3}{{{m^2}}}}}{{\frac{3}{{{m^2}}}}} = \frac{4}{3} \cr & \Rightarrow \frac{{{m^2} - 4m + 4 - 6}}{3} = \frac{4}{3} \cr & \Rightarrow {m^2} - 4m - 6 = 0 \cr} $$
$${m_1}$$ and $${m_2}$$ are roots of this equation, therefore $${m_1} + {m_2} = 4\,\,{\text{and }}{m_1}{m_2} = - 6$$
The given expression is, $$\frac{{m_1^2}}{{{m_2}}} + \frac{{m_2^2}}{{{m_1}}} = \frac{{m_1^3 + m_2^3}}{{{m_1}{m_2}}}$$
$$\eqalign{ & = \frac{{{{\left( {{m_1} + {m_2}} \right)}^3} - 3{m_1}{m_2}\left( {{m_1} + {m_2}} \right)}}{{{m_1}{m_2}}} \cr & = \frac{{{{\left( 4 \right)}^3} - 3 \cdot \left( { - 6} \right) \cdot \left( 4 \right)}}{{ - 6}} = - \frac{{68}}{3} \cr} $$

We have
$${\alpha ^2} = 5\alpha - 3\,\,{\text{and }}{\beta ^2} = 5\beta - 3;$$
$$ \Rightarrow \,\,\alpha \, \& \, \beta $$  are roots of equation, $${x^2} = 5x - 3\,$$
or $${x^2} - 5x + 3 = 0$$
$$\therefore \,\,\alpha + \beta = 5\,\,{\text{and }}\alpha \beta {\text{ = 3}}$$
Thus, the equation having $$\frac{\alpha }{\beta }\& \frac{\beta }{\alpha }$$ as its roots is
$$\eqalign{ & {x^2} - x\left( {\frac{\alpha }{\beta } + \frac{\beta }{\alpha }} \right) + \frac{{\alpha \beta }}{{\alpha \beta }} = 0 \cr & \Rightarrow \,\,{x^2} - x\left( {\frac{{{\alpha ^2} + {\beta ^2}}}{{\alpha \beta }}} \right) + 1 = 0\,\, \cr & {\text{or }}3{x^2} - 19x + 3 = 0 \cr} $$

Given equation are
$$\eqalign{ & {x^2} + 2x + 3 = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\text{i}} \right) \cr & a{x^2} + bx + c = 0\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{ii}}} \right) \cr} $$
Roots of equation (i) are imaginary roots.
According to the question (ii) will also have both roots same as (i). Thus
$$\eqalign{ & \frac{a}{1} = \frac{b}{2} = \frac{c}{3} = \lambda \,\,\,\,\,\left( {{\text{say}}} \right) \cr & \Rightarrow \,\,a = \lambda ,b = 2\lambda ,c = 3\lambda \cr} $$
Hence, required ratio is $$1 : 2 : 3$$

$$\eqalign{ & \frac{{{x^2} - 3x + 4}}{{x + 1}} > 1 \cr & \Rightarrow \frac{{{x^2} - 3x + 4}}{{x + 1}} - 1 > 0 \cr & \Rightarrow \frac{{{x^2} - 4x + 3}}{{x + 1}} > 0 \cr & \Rightarrow \frac{{\left( {x + 1} \right)\left( {x - 1} \right)\left( {x - 3} \right)}}{{{{\left( {x + 1} \right)}^2}}} > 0 \cr & \Rightarrow \left( {x + 1} \right)\left( {x - 1} \right)\left( {x - 3} \right) > 0\,\,{\text{and }}x \ne - 1 \cr} $$
Using method of interval, we get,
$$x \in \left( { - 1,1} \right) \cup \left( {3,\infty } \right)$$