Too Many Greek Letters

Note that the polynomial $F(\frac{1}{x})$ has roots $\frac{1}{\alpha}$ , $\frac{1}{\beta}$ , etc.

Therefore, we can compute $F(\frac{1}{x})$ then use Vieta's to get the sum of its roots.

We see that $F\left(\frac{1}{x}\right)=G(x)=7x^6-6x^5+5x^4-\cdots$ .

Since all we need to know is the coefficients of the first two terms, we can disregard the rest.

Our sum is then $-(-6)\div 7=\frac{6}{7}$ so our answer is $6\times 7=\boxed{42}$ .

The expression given can be rewritten carefully as:

$\frac {\alpha\beta\gamma\theta\sigma + \alpha\beta\gamma\theta\omega + \alpha\beta\gamma\sigma\omega + \alpha\beta\theta\sigma\omega + \alpha\gamma\theta\sigma\omega + \beta\gamma\theta\sigma\omega}{\alpha\beta\gamma\theta\sigma\omega}$ .

By Vieta's formulas, we know the numerator is equal to $- \frac {-6}{1} = 6$ , and the denominator is equal to $\frac {7}{1} = 7$ . Hence, the expression is equal to $\frac {6}{7}$ , and $a \times b = 6 \times 7 = \boxed {42}$ .

Finding a common denominator, we have

$\dfrac1\alpha + \dfrac1\beta + \dfrac1\gamma + \dfrac1\theta + \dfrac1\sigma + \dfrac1\omega = \dfrac{\alpha\beta\gamma\theta\sigma + \alpha\beta\gamma\theta\omega + \alpha\beta\gamma\sigma\omega + \alpha\beta\theta\sigma\omega + \alpha\gamma\theta\sigma\omega + \beta\gamma\theta\sigma\omega}{\alpha\beta\gamma\theta\sigma\omega}.$

( Way too many Greek letters!) By Vieta's Formulas, the numerator of this fraction is equal to the negative of the $x$ coefficient, divided by the $x^6$ coefficient, which is $-\left(\dfrac{-6}{1}\right) = 6,$ and the numerator of this fraction is equal to the constant divided by the $x^6$ coefficient, which is $\dfrac71 = 7.$ Thus, the value of the expression we want is $\dfrac67,$ and the answer is $6 \times 7 = \boxed{42}.$

the sum of the expression will be a fraction numerator of which is sum of the roots taken 5 at a time and denominator is product of roots. using theory of equations the numerator=-(-6)/1=6 and denominator=7/1=7 a=6 b=7 axb=42....

tire o mmc;;;;;,sendo:a,b,c,d,e,f ,g raizes::: (bcdefg+acdefg+abdefg+abcefg+abcdfg+abcdeg+abcdef)/abcdefg

por Girard tem-se:

bcdefg+acdefg+abdefg+abcefg+abcdfg+abcdeg+abcdef=-f/a=-(-6)/1=6

e

abcdefg=g/a=7/1=7

logo

(bcdefg+acdefg+abdefg+abcefg+abcdfg+abcdeg+abcdef)/abcdefg=6/7....reposta 6x7=42

the simplify is

a = $\frac{-f}{a}$ = $\frac{6}{1}$ = $\boxed{6}$

b = $\frac{g}{a}$ = $\frac{g}{a}$ = $\boxed{7}$

a $\times$ b = 6 $\times$ 7 = $\boxed{42}$

Before we see the polynomial of $6$ th degree, let's see a $3$ rd degree polynomial.

$F(x) = ax^{3} + bx^{2} + cx + d$

We know here, if its roots are $\alpha$ , $\beta$ and $\gamma$ , then

$\alpha + \beta + \gamma = \frac{-b}{a}$

$\alpha \beta + \beta \gamma + \gamma \alpha = \frac{c}{a}$

$\alpha \beta \gamma = \frac{-d}{a}$

Hence in any polynomial the $(-1)^{Degree Of Polynomial} \times \frac{Constant}{Coefficient Of Highest Power}$ represents the product of all its roots.

Here, we need to find

$\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} + \frac{1}{\theta} + \frac{1}{\sigma} + \frac{1}{\omega}$

$= \frac{Sum Of Products Of Roots (5 At A Time)}{Product Of All The Roots}$

Product of all the roots is $7$ .

Sum of Products of Roots (5 at a time) is $-(-6) = 6$

Hence, the required result is

$= \frac{6}{7}$

$a \times b = 6 \times 7 = 42$

That's the answer!

$1/\alpha+1/\beta+1/\gamma+1/\theta+1/\sigma+1/\omega = \frac{\beta*\gamma*\theta*\sigma*\omega+\alpha*\gamma*\theta*\sigma*\omega+\alpha*\beta*\theta*\sigma*\omega+\alpha*\beta*\gamma*\sigma*\omega+\alpha*\beta*\gamma*\theta*\omega+\alpha*\beta*\gamma*\theta*\sigma}{\alpha*\beta*\gamma*\theta*\sigma*\omega}$

The numerator $a$ is the sum of the products of the roots taken five at a time. By Vieta's formulas, this is equal to $-6$ the negative of the second to last coefficient. The denominator $b$ is equal to the product of the roots, which is $7$ , the constant term of the function. $a*b=6*7=\fbox{42}$

The polynomial does not have any rational roots, so we need to see if we can use an application of Vieta's formula to solve for the unknown expression. This may be useful.

If a polynomial $F(x)=c_1x^n+c_2x^{n-1}+\ldots+c_nx+c_{n+1}$ has roots $r_1\text{,}$ $r_2\text{,}$ $\ldots\text{,}$ $r_n\text{,}$ then the polynomial with roots $\frac{1}{r_1}\text{,}$ $\frac{1}{r_2}\text{,}$ $\ldots\text{,}$ $\frac{1}{r_n}$ can be represented by $F^*(x)=c_{n+1}x^n+c_nx^{n-1}+\ldots+c_2x+c_1$ .

Using this, we can determine that if $F(x)=x^6-2x^5+3x^4-4x^3+5x^2-6x+7$ has roots $\alpha\text{,}$ $\beta\text{,}$ $\gamma\text{,}$ $\theta\text{,}$ $\sigma\text{,}$ and $\omega\text{,}$ then the polynomial with roots $\frac{1}{\alpha}\text{,}$ $\frac{1}{\beta}\text{,}$ $\frac{1}{\gamma}\text{,}$ $\frac{1}{\theta}\text{,}$ $\frac{1}{\sigma}\text{,}$ and $\frac{1}{\omega}$ is $F^*(x)=7x^6-6x^5+5x^4-4x^3+3x^2-2x+1$ .

The sum of the roots of a polynomial $Ax^n+Bx^{n-1}+\ldots$ is $\frac{-B}{A}$ . In $F^*(x)\text{,}$ $A=7$ and $B=-6$ . The sum of the roots of $F^*(x)$ is $\frac{-(-6)}{7}=\frac{6}{7}$ . For the problem, $a=6$ and $b=7$ , so $a\times b=\boxed{42}$ .

The expression of which we need to find the value can be rewritten by taking L.C.M. as,

       sum(roots taken 5 at a time)/ product(the six roots)

Considering Vieta's Formula for a general equation of degree 'n', we can simplify the expression as,

     sum(roots taken 5 at a time) = (-1)^5 * coefficient of x^(6-5)  /  coefficient of x^6 
                                                             = -(-6)/ 1 = 6   

     product of the six roots = (-1)^6 * coefficient of x^(6-6) / coefficient of x^6 = 7

Thus, a/b = 6/7 ===> aXb = 6X7 = 42...

With Vietha's Theorem that "if a,b,c,d,e,f are roots of the polynomial above, So that bcdef + acdef + abdef + abcef + abcdf + abcde = - (-6)/1 = 6 and abcdef = + (7)/1 = 7. Thius, a= 6 and b = 7. So, ab = 6.7 = 42 (Answer)