×

The must be a faster way - 1

If $$p,q,r$$ are the roots of the equation

$x^3-3px^2+3q^2x-r^3$

Prove that $$p=q=r$$.

Elementary proof:

$$p+q+r=3p$$

$$pq+qr+rp=3q^2$$

$$pqr=r^3$$

From the third equation,

$$pq=r^2$$

Substitute to the second equation,

$$r^2+qr+rp=3q^2$$

$$r(p+q+r)=3q^2$$

$$3rp=3q^2$$

$$pr=q^2$$

Substitute $$pr=q^2$$ into the third equation,

$$q^3=r^3\implies q=r$$

Again,

$$pq=r^2$$

$$pr=r^2$$

$$p=r \implies p=q=r$$

Can you find a better proof?

Note by Christopher Boo
2 years, 9 months ago

Sort by:

Put $$x=r$$ in the equation. Since $$r$$ is a root, we immediately get $$pr=q^2$$ (assuming $$r\neq 0$$). Also $$pqr=r^3$$. The rest follows from these two. · 2 years, 9 months ago

Brilliant proof!!! · 2 years, 9 months ago

Brilliant! @Abhishek Sinha · 2 years, 9 months ago

we can assume that p=q . Which means that we are assuming p to be the repeated root of the given function . Differentiate the given cubic and let it be g(x). then substitute p in g(x) Since we have assumed p to be repeated root thus it will also be the root of g(x). on substituting p in the equation we will get p=q which concurs with our assumption. · 2 years, 6 months ago

Indeed, I've found a much faster solution. Using the cubic formula, we get that

\begin{align*}x&=\sqrt[3]{\left(\frac{-(-3p)^3}{27(1)^3}+\frac{(-3p)(3q^2)}{6(1)^2}-\frac{(-r^3)}{2(1)}\right)+\sqrt{\left(\frac{-(-3p)^3}{27(1)^3}+\frac{(-3p)(3q^2)}{6(1)^2}-\frac{(-r^3)}{2(1)}\right)^2+\left(\frac{(3q^2)}{3(1)}-\frac{(-3p)^2}{9(1)^2}\right)^3}}\\&+\sqrt[3]{\left(\frac{-(-3p)^3}{27(1)^3}+\frac{(-3p)(3q^2)}{6(1)^2}-\frac{(-r^3)}{2(1)}\right)-\sqrt{\left(\frac{-(-3p)^3}{27(1)^3}+\frac{(-3p)(3q^2)}{6(1)^2}-\frac{(-r^3)}{2(1)}\right)^2+\left(\frac{(3q^2)}{3(1)}-\frac{(-3p)^2}{9(1)^2}\right)^3}}\\&-\frac{(-3p)}{3(1)}\\&=\sqrt[3]{p^3-\frac{3pq^2}2+\frac{r^3}2+\sqrt{\left(p^3-\frac{3pq^2}2+\frac{r^3}2\right)^2+\left(q^2-p^2\right)^3}}\\&+\sqrt[3]{p^3-\frac{3pq^2}2+\frac{r^3}2-\sqrt{\left(p^3-\frac{3pq^2}2+\frac{r^3}2\right)^2+\left(q^2-p^2\right)^3}}\\&+p\end{align*}

is one of $$p$$, $$q$$, and $$r$$. We can find the other roots by dividing out, and the rest of the proof is omitted. · 2 years, 9 months ago