Product of Roots of Sums of Products of Differences

This equation will be formed in $ax^3+bx^2+cx+d=0$ . This equation has 3 roots let's say $p,q,r$ .

Using Vieta theorem we will get $p+q+r = -b/a, pq+qr+pr = c/a, pqr = -d/a$ . We are just asked to find the value of $pqr$ , so need to determine $d$ and $a$ . Because $d$ is a constant so we'll get $d$ equals to $(-1)(-2)(-3) + (-2)(-3)(-4) + (-3)(-4)(-5) + (-4)(-5)(-6) + (-5)(-6)(-7) + (-6)(-7)(-8) = -756$ . Because $a$ is the coefficient of $x^3$ so we'll get $a$ equals to $1+1+1+1+1+1 = 6$ . So we get $pqr = -d/a = -(-756)/6 = 126$ . So the product of all roots to the equation is 126.

(x-a)(x-b)(x-c) = x³ – (a+b+c)x² + (ab+bc+ca)x – abc

We can express all the six polynomials in this manner and add them up to get:

6x³ - $\displaystyle \sum_{i=1}^6$ ( $a_i + b_i + c_i$ )x² + $\displaystyle \sum_{i=1}^6$ ( $a_i.b_i + b_i.c_i + c_i.a_i$ )x - $\displaystyle \sum_{i=1}^6$ ( $a_i . b_i . c_i$ ) = 0

Now we know that the product of the roots of an cubic polynomial ax³ + bx² + cx + d = 0 can be given as $\frac {-d}{a}$

Using this knowledge we can see that the product of the roots of the above polynomial can be expressed as $\frac {(\displaystyle \sum_{i=1}^6)((a_i.b_i.c_i)}{6}$ .

Here by calculation we get $\displaystyle \sum_{i=1}^6$ ( $a_i.b_i.c_i$ ) as 756 .

Hence the product of the roots is 126

The equation can be rewritten in the form: $3(2x-9)(x^2-9x+28)=0$ which has one real root $x=\frac{9}{2}$ . The other 2 roots of the equation $x^2-9x+28=0$ have the product equal to 28 (by Viete). So the product of all roots is 126.

Suppose that the L.H.S. of the given equation can be expand as $a_3x^3+a_2x^2+a_1x+a_0$ . By using Vieta's formulas, we obtain that the product $P$ of all roots to the given equation is equal to $-\frac{a_0}{a_3}$ . It is easy to see that $a_3 = 6$ and $a_0 = \sum_{i=1}^6 i(i+1)(i+2) = -756$ . Therefore $P = \frac{756}{6}=126$ .

each term is a cubic expression of the form (x-a)(x-b)(x-c). The roots of each expression are a,b and c. For example the first term is (x-1)(x-2)(x-3), so its roots are 1,2 and 3. The product of roots of a cubic equation is given by -(coefficient of x^0)/(coefficient of x^3). Here, coefficient of x^3 is 1. And product of roots is 1 2 3=6. So the cubic expression of which these are roots is given by x^3+.........-6. Similarly the constants of the other five cubic expressions can be found in a similar way. They are -24,-60,-120,-210 and -336, respectively. So the combined constant of the given equation is the sum of all these constants i.e. -6+-24+-60+-120+-210+-336=-756. So the resulting cubic expression has a constant term of -756. However, the coefficient of x^3 in this case is 6, as the equation is composed of six cubic expressions. So, product of roots=-(-756)/6=126.

Expand the given expression completely to get a cubic equation :

2x^3 - 27x^2 + 137x - 252 = 0.

In this case we get the product of all three roots as

-constant term / (coefficient of x^3). That gives 252 / 2 = 126.

In the given question, we will observe that on opening all the brackets the given expression will take the form - ax^3 + bx^2 + cx^1 + d = 0

For such cubic expressions :- Product of the roots is given by -d/a

So to find the given result we will just have to get the ( )x^3 and the constant term

By simple observation we will get the result as

6x^3 + ( unwanted term )x^2 + ( unwanted term )x^1 + ( constant - d)

constant term is -{ 1 2 3 + 2 3 4 + 3 4 5 + 4 5 6 + 5 6 7 + 6 7 8 } constant term = d= -756

So product of roots is -(-756)/6 = 756/6 = 126

The equation in the question above can be written as $\begin{aligned} 6x^3-81x^2+411x-756 &=0 \\ 3(2x-9)(x^2-9x+28) &=0 \\ (2x-9)(x^2-9x+28) &=0 \end{aligned}$ Then we have $\begin{aligned} 2x-9 &=0\\ x_1 &= \frac{9}{2} \end{aligned}$ and $\begin{aligned} x^2-9x+28 &=0 \\ x^2-9x &=-28\\ \left(x-\frac{9}{2}\right)^2 -\frac{81}{4}&=-28\\ \left(x-\frac{9}{2}\right)^2 &=-28+\frac{81}{4}\\ x-\frac{9}{2} &= \sqrt{-\frac{31}{4}}\\ x-\frac{9}{2} &= \pm \frac{1}{2}i \sqrt{31}\\ x &= \frac{1}{2} (9 \pm i \sqrt{31})\\ x_2 &= \frac{1}{2} (9 + i \sqrt{31}) \text{ ; } x_3 = \frac{1}{2} (9 - i \sqrt{31}) \end{aligned}$ Finally, $\begin{aligned} x_1 \cdot x_2 \cdot x_3 &= \frac{9}{2} \cdot \frac{1}{2} (9 + i \sqrt{31}) \cdot \frac{1}{2} (9 - i \sqrt{31})\\ &= \frac{9}{8}(9^2 - (i \sqrt{31})^2)\\ &= \frac{9}{8}(81 + 31)\\ &= \boxed{126} \end{aligned}$

$\text{\# }\mathbb{Q}.\mathbb{E}.\mathbb{D}.\text{\#}$

If $p(x) = ax^3 + bx^2 + cx + d$ , then the product of the roots is $-d/a$ .

To find $d$ we may evaluate $p(0) = (-1\cdot-2\cdot-3 )+(-2\cdot-3\cdot-4)+\cdots+(-6\cdot-7\cdot-8) = -3!(\binom3 3 +\binom 4 3 +\binom 5 3+\cdots+\binom 8 3) = -3!(\binom9 4) = -756$ .

To find $a$ we may consider $p(\dfrac1 x)$ : $(\dfrac1 x -1)(\dfrac1 x -2)(\dfrac1 x -3) +\cdots+(\dfrac1 x -6)(\dfrac1 x - 7)(\dfrac1 x -8) = (1-x)(1-2x)(1-3x)+\cdots+(1-6x)(1-7x)(1.8x)$ . Evaluating in $0$ gives us $a=6$ . Hence the product is $-(-756)/(6)=126$ .

After expanding i got : 6x^3-81x^2+421=756 and we know that product of roots is constant term divided by coefficient of term with highest power i.e 756/6=126