# 2002 Indonesia MO Problems/Problem 3

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## 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 byProblem 2 1 • 2 • 3 • 4 • 5 • 6 • 7 Followed byProblem 4 All Indonesia MO Problems and Solutions
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