If p,q,r are the roots of the equation
x3−3px2+3q2x−r3
Prove that p=q=r.
Elementary proof:
By Vieta's Formula,
p+q+r=3p
pq+qr+rp=3q2
pqr=r3
From the third equation,
pq=r2
Substitute to the second equation,
r2+qr+rp=3q2
r(p+q+r)=3q2
3rp=3q2
pr=q2
Substitute pr=q2 into the third equation,
q3=r3⟹q=r
Again,
pq=r2
pr=r2
p=r⟹p=q=r
Can you find a better proof?
#Algebra
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Put x=r in the equation. Since r is a root, we immediately get pr=q2 (assuming r=0). Also pqr=r3. The rest follows from these two.
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Brilliant proof!!!
Brilliant! @Abhishek Sinha
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.
Indeed, I've found a much faster solution. Using the cubic formula, we get that
x=3(27(1)3−(−3p)3+6(1)2(−3p)(3q2)−2(1)(−r3))+(27(1)3−(−3p)3+6(1)2(−3p)(3q2)−2(1)(−r3))2+(3(1)(3q2)−9(1)2(−3p)2)3+3(27(1)3−(−3p)3+6(1)2(−3p)(3q2)−2(1)(−r3))−(27(1)3−(−3p)3+6(1)2(−3p)(3q2)−2(1)(−r3))2+(3(1)(3q2)−9(1)2(−3p)2)3−3(1)(−3p)=3p3−23pq2+2r3+(p3−23pq2+2r3)2+(q2−p2)3+3p3−23pq2+2r3−(p3−23pq2+2r3)2+(q2−p2)3+p
is one of p, q, and r. We can find the other roots by dividing out, and the rest of the proof is omitted.
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Honestly, why do you love to bash out all the problems so much?