The must be a faster way - 1

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

x33px2+3q2xr3x^3-3px^2+3q^2x-r^3

Prove that p=q=rp=q=r.


Elementary proof:

By Vieta's Formula,

p+q+r=3pp+q+r=3p

pq+qr+rp=3q2pq+qr+rp=3q^2

pqr=r3pqr=r^3

From the third equation,

pq=r2pq=r^2

Substitute to the second equation,

r2+qr+rp=3q2r^2+qr+rp=3q^2

r(p+q+r)=3q2r(p+q+r)=3q^2

3rp=3q23rp=3q^2

pr=q2pr=q^2

Substitute pr=q2pr=q^2 into the third equation,

q3=r3    q=rq^3=r^3\implies q=r

Again,

pq=r2pq=r^2

pr=r2pr=r^2

p=r    p=q=rp=r \implies p=q=r


Can you find a better proof?

Note by Christopher Boo
5 years ago

No vote yet
1 vote

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Put x=rx=r in the equation. Since rr is a root, we immediately get pr=q2pr=q^2 (assuming r0r\neq 0). Also pqr=r3pqr=r^3. The rest follows from these two.

Abhishek Sinha - 5 years ago

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Brilliant proof!!!

Anuj Shikarkhane - 5 years ago

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Brilliant! @Abhishek Sinha

Christopher Boo - 5 years 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 - 4 years, 9 months ago

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

x=((3p)327(1)3+(3p)(3q2)6(1)2(r3)2(1))+((3p)327(1)3+(3p)(3q2)6(1)2(r3)2(1))2+((3q2)3(1)(3p)29(1)2)33+((3p)327(1)3+(3p)(3q2)6(1)2(r3)2(1))((3p)327(1)3+(3p)(3q2)6(1)2(r3)2(1))2+((3q2)3(1)(3p)29(1)2)33(3p)3(1)=p33pq22+r32+(p33pq22+r32)2+(q2p2)33+p33pq22+r32(p33pq22+r32)2+(q2p2)33+p\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 pp, qq, and rr. We can find the other roots by dividing out, and the rest of the proof is omitted.

Cody Johnson - 5 years ago

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

Sagnik Saha - 5 years ago

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