Appropriate Substitution

First method - direct and easy method

$( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} )^{2}$

$= ( \frac{ bc + ac + ab }{ abc } )^{2}$

Now,

We know that,

$abc = -8 \\ ab + bc + ac = 12$

Therefore,

$= ( \frac{ bc + ac + ab }{ abc } )^{2}$

$=( \frac{12}{8})^{2}= \frac{m}{ n}$

$=( \frac{3}{2})^{2} = \frac{m}{n}$

Therefore,

$16(m - n) = \boxed{80}$

Second method - a bit concept based one

The equation who's roots will be $\frac{1}{a} , \frac{1}{b} , \frac{1}{c}$ is,

$8x^{3} - 12x^{2} + 6x - 1 = 0$

Now,

Sum of roots = $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{12}{8}$

$= \frac{3}{2}$

Therefore,

$\frac{m}{n} = ( \frac{3}{2} )^{2} = \frac{9}{4}$

Therefore,

$16( m - n ) = \boxed{80}$

$\text {The given equation}$ $x^{3} - 6x^{2} + 12x = 8$ $\text {is simply the expansion of}$ $(x-2)^{3}=0$ .

$\text {hence,}$ $a=b=c=2$

$So,$ $( \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} )^{2} = \dfrac{m}{n} = \dfrac{3^2}{2^2}= \dfrac{9}{4}$

$\text {Therefore,}$ $16(m-n) = 16\times (9-4) = 80$

$\text {I would have used the same method as Sakanksha Deo}$ $\text {if the equation would have been a bit different}$ $\text {but the given equation is a simple one therefore not proceeded that way.}$

a,b,c are the roots From the eqn ab+bc+ac=12,abc=-8 (1/a + 1/b + 1/c)^2=((ab+bc+ac+)/abc)^2=(12/-8)^2=(-3/2)^2=9/4 9/4=m/n m=9,n=4 16(m-n)=16(9-5)=16(5)=80

x^3 - 6x^2 + 12x - 8 = 0 ; (x-2)^3 = 0 ; x = 2 ; all the three root are equal and its 2 ; so a = 2; b = 2; c = 2 ; (1/a + 1/b + 1/c)^2 = 9/4 = m/n ; m = 9; n = 4; 16(9-4) = 80

or else replace x by 1/x :P so that the roots are multiplicative inverse of themselves ..!:P