Difference between revisions of "2002 Indonesia MO Problems/Problem 3"

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==See Also==
 
==See Also==
{{Indonesia MO 7p box|year=2002|num-b=2|num-a=4}}
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{{Indonesia MO box|year=2002|num-b=2|num-a=4|eight=}}
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]

Latest revision as of 00:09, 4 August 2018

Problem

Find all real solutions from the following system of equations:

$\left\{\begin{array}{l}x+y+z = 6\\x^2 + y^2 + z^2 = 12\\x^3 + y^3 + z^3 = 24\end{array}\right.$

Solution

Square the first equation to get \[x^2 + y^2 + z^2 + 2(xy + yz + xz) = 36\] Subtract the second equation from the result to get \[2(xy+yz+xz) = 24\] \[xy+yz+xz = 12\] Multiply the second equation by the first equation to get \[x^3 + xy^2 + xz^2 + x^2y + y^3 + yz^2 + x^2z + y^2z + z^3 = 72\] Subtract the third equation to get \[xy^2 + xz^2 + x^2y + yz^2 + x^2z + y^2z = 48\] Cube the first equation to get \[(x^3 + y^3 + z^3) + 3(x^2y + x^2z + xy^2 + y^2z + xz^2 + yz^2) + 6xyz = 216\] \[24 + 3(48) + 6xyz = 216\] \[168 + 6xyz = 216\] \[6xyz = 48\] \[xyz = 8\] If $x+y+z=6$, $xy+yz+xz = 12$, and $xyz = 8$, the solution triplet is the roots of the polynomial \[a^3 - 6a^2 + 12a - 8 = 0\] Factor the polynomial to get \[(a-2)^3 = 0\] Since $a = 2$ is a triple root to the polynomial, the only solution to the system of equations is $\boxed{(2,2,2)}$, and plugging the values back in satisfies the system.

See Also

2002 Indonesia MO (Problems)
Preceded by
Problem 2
1 2 3 4 5 6 7 Followed by
Problem 4
All Indonesia MO Problems and Solutions
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