Rational solution to $X^3 -6X -6 = 0$

Question

So I have to prove that there is no rational solution to the Polynomial (I dont know if I spelled it correctly it's Polynôme in french) $ X^3 - 6X -6 = 0 $ I know that i have to suppose that there exists a rational solution for the polynomial and then find a contradiction but I tried almost every trick up my sleeve and I haven't been able to prove it. I thought that I could demonstrate by supposing that a/b (two integers) is the rational solution Gcd(a,b) =1 and then finding a common divisor [I don't know the English terms very well] but i just can't progress using this classical method. I'm in terminal highschool if you want to know my level. P.s: thanks in advance.

Answer 1

Suppose the equation $x^3 - 6x -6 = 0$ has a rational root ${\large{\frac{p}{q}}}$, where $p,q$ are integers with $\gcd(p,q)=1$. \begin{align*} \text{Then}&\;\;x^3-6x-6=0 \qquad\qquad\;\\[4pt] \implies\;&\left({\small{\frac{p}{q}}}\right)^3-6\left({\small{\frac{p}{q}}}\right)-6=0\\[4pt] \implies\;&p^3-6pq^2-6q^3=0\\[4pt] \implies\;&p^3=2(3pq^2+3q^3)\\[4pt] \end{align*} hence $p$ must be even, so we can write $p=2s$, for some integer $s$. \begin{align*} \text{Then}&\;\;p^3-6pq^2-6q^3=0\\[4pt] \implies\;\;&(2s)^3-6(2s)q^2-6q^3=0\\[4pt] \implies\;\;&8s^3-12sq^2-6q^3=0\\[4pt] \implies\;\;&3q^3=2(2s^3-3sq^2)\\[4pt] \end{align*} hence $q$ must be even, contradiction, since $\gcd(p,q)=1$.

Thus, the given equation has no rational roots.

Answer 2

There's a rational roots theorem, which may no be in your curriculum, in which case you can prove it, then use it:

Let $f(x)=c_nx^n+c_{n-1}x^{n-1}+\dots+c_1x+c_0\enspace(c_n\ne 0)\;$ be a polynomial with integer coefficients. If $a/b$ is a rational root in irreducible form, then $a$ divides the constant term $c_0$ and $b$ divides the leading coefficient $c_n$.

Answer 3

There are various ways of solving this, including using general results. Here is a direct method. Let $X=\frac ab$ be a fraction in lowest terms, with $b\gt 0$. Then $a^3-6ab^2-6b^3=0$ so that $$a^3=6b^2(a+b)$$ from which it follows that $a$ is divisible by both $2$ and $3$ as well as any prime factor of $b$. But we assumed that $a$ and $b$ had no common factors, so $b$ is positive and has no prime factors, so we must have $b=1$.

[Note that this argument can be carried out for any polynomial where the coefficient of the highest power of $x$ is equal to $1$ - you can pull out a factor of $b$ from the lower terms and get that a multiple of $b$ is equal to a power of $a$. In such a case any rational root is an integer. Then we can use the factor theorem for polynomials to show that any integer root divides the constant term - which is the substance of the rational root test applied to monic polynomials - it is not a scary thing]

So we can go back to the original equation and simply assume now that $X=a$ is an integer. Then $$X^3=6(X+1)$$

$X$ must be even, because the right-hand side is, but $X+1$ is odd. $X^3$ is divisible by $8$ and $6(x+1)$ is not.

Answer 4

Another way is to solve the equation and find the actual root.

To prove that the equation has exactly 1 root, notice that by differentiating the function f(x) = x^3 - 6x - 6, you'll know where the local maximum and minimum are. f'(x) = 3x^2 - 6 = 0 eq. x = +/- 2^(1/2). The negative value is at a local maximum, the positive value at a local minimum, since f is positive for x >> 0. You can determine that f(-2^(1/2)) < 0 and f(2^(1/2)) > 0. This leads to the conclusion that the equation has only 1 real root.

Applying a solution method, you can find that the only root is equal to 2^(1/3) + 4^(1/3), but you can check this without ado. This root is clearly not rational.