Polynomial Vieta

By Vieta's Formula ,

$\alpha+\beta+\gamma=-\frac{2}{1}=-2$

$\alpha\beta+\alpha\gamma+\beta\gamma=\frac{3}{1}=3$

$\alpha\beta\gamma=-\frac{4}{1}=-4$

Hence,

$\frac{\alpha+\beta}{\gamma}+\frac{\beta+\gamma}{\alpha}+\frac{\gamma+\alpha}{\beta}$

$=\frac{-2-\gamma}{\gamma}+\frac{-2-\alpha}{\alpha}+\frac{-2-\beta}{\beta}$

$=-\frac{2}{\gamma}-1-\frac{2}{\alpha}-1-\frac{2}{\beta}-1$

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

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

$=-(\frac{2(\alpha\beta+\alpha\gamma+\beta\gamma)}{-4})-3$

$=\frac{2(3)}{4}-3=\boxed{-\frac{3}{2}}$

Subtracting and adding $3$ to both sides of the given expression yields:

$\frac{\alpha + \beta + \gamma}{\gamma} + \frac{\alpha + \beta + \gamma}{\alpha} + \frac{\alpha + \beta + \gamma}{\beta} - 3$

$=(\alpha+\beta+\gamma)\left(\frac{1}{\gamma} + \frac{1}{\alpha} + \frac{1}{\beta}\right)-3$

$= (\alpha + \beta+\gamma)\left(\frac{\alpha\beta +\beta\gamma+\gamma\alpha }{\beta\gamma\alpha} \right)-3$

We can easily calculate the values of these cyclic sums using Vieta's formula from the given polynomials, substitute the values, and get the desired answer.

Prepare the equations we need first so that it will be easier later when we try to solve it.

By Vieta's Formula,

$\alpha+\beta+\gamma=-2$

$\alpha\beta+\beta\gamma+\gamma\alpha=3$

$(\alpha\beta\gamma)=-4$

Now that we've prepared everything necessary, we start by turning the problem given into a single fraction and see what we can do next

$\frac{\alpha+\beta}{\gamma}+\frac{\beta+\gamma}{\alpha}+\frac{\gamma+\alpha}{\beta}$

$=\frac{\alpha\beta(\alpha+\beta)+\beta\gamma(\beta+\gamma)+\gamma\alpha(\alpha+\gamma)}{\alpha\beta\gamma}$

We still can't do anything with this, we need to turn this fraction into something we can solve with the equations prepared earlier.

So, we can let $\alpha+\beta$ become $\alpha+\beta+\gamma-\gamma$ and the same thing with the other ones.

$\frac{\alpha\beta(\alpha+\beta+\gamma-\gamma)+\beta\gamma(\alpha+\beta+\gamma-\alpha)+\gamma\alpha(\alpha+\beta+\gamma-\beta)}{\alpha\beta\gamma}$

Now we can almost solve it by replacing $\alpha+\beta+\gamma$ with $-2$ , but there are still some that we can't substitute a value into it.

We can multiply out the extra negative value from the bracket.

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

Then,

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

Finally, substitute in the values we'd precomputed

$\frac{3(-2)-3(-4)}{-4}$

$=\frac{6}{-4}$

$=\boxed{-\frac{3}{2}}$

Sum of all roots = -2; sum of two roots at a time = 3; product of roots = -4

Combining the expression, you get ((a + b + c)(ab + bc + ac) - 3abc)/abc = -3/2

We see that $(a+b+c)(ab+bc+ca)=3abc+\sum a^2 b$ and the fraction we want is $\frac{\sum a^2 b}{abc}$ . Hence, we plug in and finish with Vieta.

from given polynomial equation........ α+β+ϒ=-2 αβ + βϒ+ βϒ=3 αβϒ=-4 now, (α+β)/ϒ+(β+ϒ)/α+(ϒ+α)/β

=> (-2-ϒ)/ϒ + (-2-α)/α + (-2-β)/β

=> ((-2)/ϒ + (-2)/α + (-2)/β) - 3

=> ((-2(αβ + βϒ+ βϒ))/αβϒ) - 3 => ((-2×3)/(-4))-3= - 3/2

a + b + c = -2.....(a + b)/c = ( -2 - c )/c = -2/c - 1...(b+c)/a = - 2/a -1...(c+a)/b = -2/b - 1

Exp. = -2/c -1 - 2/b - 1 - 2/a -1 = -2{ 1/c +1/b + 1/a} - 3
Sum of reciprocals = - 3/4...
Exp. - 2 * (- 3/4 ) -3 = - 3/2