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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:

By Vieta's Formula,

\(p+q+r=3p\)

\(pq+qr+rp=3q^2\)

\(pqr=r^3\)

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

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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. Abhishek Sinha · 2 years, 9 months ago

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

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@Abhishek Sinha Brilliant! @Abhishek Sinha Christopher Boo · 2 years, 9 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. Prince Kumar Maurya · 2 years, 6 months ago

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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. Cody Johnson · 2 years, 9 months ago

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@Cody Johnson Honestly, why do you love to bash out all the problems so much? Sagnik Saha · 2 years, 9 months ago

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