The sum of the cubes of the reciprocal values of the roots of the equation $x^2+ax+1=0$ is?

Question

The equation : $$x^2+ax+1=0$$

$a$ is a real number.

How to find the sum of cubes of the reciprocal of roots for this equation.

I tried solving this just by brute forcing it, but I get expressions that are ugly and pretty sure would yield nothing in the long run. So there is probably a trick to doing this, as the possible solutions I have nice expressions.

Any hints would also be fine.

What do you know about the sum and product of the roots of that equation? — Daniel Fischer, May 27 '15 at 20:26
What am I supposed to know? I don't know of a generalized formula for the sum of roots. This is cubed reciprocal roots, so that make such a formula probably useless. — John Doe, May 27 '15 at 20:28
Forget about that for the moment. Just a good old quadratic polynomial. How are the roots and the coefficients related? — Daniel Fischer, May 27 '15 at 20:30
$$\frac1{x^3}+\frac1{y^3}=\frac{(x+y)^3-3(xy)(x+y)}{(xy)^3}$$ — Did, May 27 '15 at 20:32
Yeah, thanks Daniel :D. Sometimes your brain just doesn't work! It's $3a-a^3$. Use Vieta's formulas, for anyone wondering. — John Doe, May 27 '15 at 20:33
The reciprocal are roots of the equation $X^2+aX+1$ and $m^3 + n^3 = (m+n)^3-3mn(m+n)$ hence you answer is $-a^3+3a$ — Piquito, May 27 '15 at 20:34
Raising the usual root formula to the third? Perhaps not difficult but definitely tedious. Vieta's make it a breeze. — John Doe, May 27 '15 at 20:38

score 8 · Accepted Answer · answered May 27 '15 at 20:32

8

Hint: Let $\alpha$ and $\beta$ be the roots. Then $${1\over \alpha ^3}+{1\over \beta ^3}={\alpha^3+\beta^3 \over{\alpha^3\beta^3}} ={(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta) \over{(\alpha\beta})^3} $$

answered May 27 '15 at 20:32

Fermat

5,230

score 3 · Answer 2 · answered May 27 '15 at 20:34

According to Viete's formulas

$x_1+x_2=-a$

$x_1x_2=1$

We are looking for $\frac{1}{x_1^3}+\frac{1}{x_2^3}$

But because $x_1x_2=1$:

$\frac{1}{x_1^3}+\frac{1}{x_2^3}=\frac{x_2^3+x_1^3}{x_1^3 x_2^3}=x_2^3+x_1^3=(x_1+x_2)^3-3x1x_2^2-3x1^2x_2=(x_1+x_2)^3-3x_1-3x_2$

Now we substitute $x_1+x_2=-a$:

$(x_1+x_2)^3-3x_1-3x_2=-a^3+3a$

score 2 · Answer 3 · edited May 28 '15 at 10:02

Brute forcing, though by no means the best way, is quite feasible here. After all, this is just a quadratic equation. The roots are $\frac{-a\pm\sqrt{a^2-4}}{2}$.

The reciprocals are $\frac{-a\mp\sqrt{a^2-4}}{2}$. This can be seen by flipping the roots over and rationalizing the denominator. Or more simply let $t=1/x$. Substitute in the equation. We get $t^2+at+1=0$.

Now we need to calculate $$\left( \frac{-a+\sqrt{a^2-4}}{2} \right)^3+\left( \frac{-a-\sqrt{a^2-4}}{2} \right)^3.$$ Expand, using the ordinary expansion of $(s+t)^3$. There is pleasant cancellation, and we get $$\frac{2}{8}\left(-a^3+3(-a)(a^2-4) \right),$$ which simplifies to $3a-a^3$.

score 1 · Answer 4 · answered May 28 '15 at 10:09

Let $y=\dfrac1{x^3}$

As $x^2+ax+1=0\iff x^2+ax=-1$

Cubing both sides, $(x^2+ax)^3=(-1)^3\iff(x^3)^2+a^3(x^3)+3a(x^3)(-1)=-1$

$\implies\left(\dfrac1y\right)^2+(a^3-3a)\left(\dfrac1y\right)=-1$

$\implies y^2+(a^3-3a)y+1=0$

$\implies y_1+y_2=-\dfrac{a^3-3a}1$

The sum of the cubes of the reciprocal values of the roots of the equation $x^2+ax+1=0$ is?

4 Answers4