Roots Relation

Algebra Level 5

$\large 24x^4 - 15x^3 + 1$

Let $a,b,c,d$ be the roots of the polynomial above. If $(abc)^3+(abd)^3+(acd)^3+(bcd)^3 = \frac pq$ for coprime positive integers $p,q$ , find the value of $p+q$ .

Image Credit: Wikimedia William Anthony Granville .

The answer is 1541.

1 solution

Danish Ahmed
Jun 18, 2015

A polynomial $g(x)$ with roots $\dfrac{1}{a}$ , $\dfrac{1}{b}$ , $\dfrac{1}{c}$ and $\dfrac{1}{d}$ will be $g(x)=x^4-15x+24$ .

Using Newton Sums, we get that $\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}+\dfrac{1}{d^3}\right)-45=0$ , or

$\dfrac{\left(abc\right)^3+\left(abd\right)^3+\left(acd\right)^3+\left(bcd\right)^3}{a^3b^3c^3d^3}=45$ .

In addition, from Vieta's Formulae on $f(x)$ , we get that $abcd=\dfrac{1}{24}$ , so $a^3b^3c^3d^3=\dfrac{1}{13824}$ .

Plugging this in, we get that

$\left(abc\right)^3+\left(abd\right)^3+\left(acd\right)^3+\left(bcd\right)^3=\dfrac{45}{13824}=\dfrac{5}{1536}$ .

Can you guide me through the logic of your first statement?

Ryan Tamburrino - 5 years, 11 months ago

Let a,b,c,d be the roots of equation 24x^4 -15x^3 + 1 = 0. Then x=a,b,c,d. Therefore if I want the equation with roots 1/a ,1/b,1/c,1/d replace x by 1/x (x=a or 1/x = 1/a) . Therefore the new equation will be x^4 - 15 + 1=0.

Rutvik Doshi - 5 years, 11 months ago

Roots Relation

Image Credit: Wikimedia William Anthony Granville .

The answer is 1541.

1 solution

0 pending reports