# 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
6 years, 8 months ago

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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.

- 6 years, 5 months ago

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

\begin{aligned}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{aligned}

is one of $p$, $q$, and $r$. We can find the other roots by dividing out, and the rest of the proof is omitted.

- 6 years, 8 months ago

Honestly, why do you love to bash out all the problems so much?

- 6 years, 8 months ago

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.

- 6 years, 8 months ago

Brilliant proof!!!

- 6 years, 8 months ago

Brilliant! @Abhishek Sinha

- 6 years, 8 months ago