Let $a+b+c=12$, $a^2+b^2+c^2=50$, and $a^3+b^3+c^3=168$. Find $a,b,c$
Suppose $a, b, c$ are roots of $P(x)$.
$$P(x) = k(x - a)(x - b)(x - c)$$
But then I get $(k = 1)$
$$P(x) = x^3 - 12x^2 + x(ab + ac + bc) - abc$$
Cant go further...
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Martin Sleziak
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Amad27
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Do you want integral solution ? – Empty May 12 '15 at 08:42
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If yes then check the number $168$.. – Empty May 12 '15 at 08:45
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1Similar questions: http://math.stackexchange.com/questions/27394/three-variable-system-of-simultaneous-equations/ or http://math.stackexchange.com/questions/403454/is-there-a-simpler-approach-to-these-system-of-equations (You could probably find more questions like this.) – Martin Sleziak May 12 '15 at 08:51
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The point is whether you can express $a^2+b^2+c^2$ and $a^3+b^3+c^3$ using $a+b+c$, $ab+ac+bc$ and $abc$. See Newton's identities at Wikipedia. – Martin Sleziak May 12 '15 at 08:53
2 Answers
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\begin{eqnarray}ab+bc+ca=\frac{1}{2}((a+b+c)^2-(a^2+b^2+c^2))=47\end{eqnarray} \begin{align*}a^3+b^3+c^3-3abc&=(a+b+c)(a^2+b^2+c^2-(ab+bc+ca))\\ abc&=44\end{align*} So $a, b, c$ are roots of the polynomial $x^3-12x^2+47x-44=0$.
Alex Fok
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$$a^2+b^2+c^2=(a+b+c)^2-2(ab+ac+bc)$$
So $ab+ac+bc=\frac{(a+b+c)^2-(a^2+b^2+c^2)}{2}=47$
Similarly
$$a^3+b^3+c^3=(a+b+c)^3+3abc-3(a+b+c)(ab+ac+bc)$$
So $abc=\frac{a^3+b^3+c^3+3(a+b+c)(ab+ac+bc)-(a+b+c)^3}{3}=44$ And $a,b,c$ are solutions of the cubic
$$X^3-12X^2+47X-44=0$$
No simple solution except Cardan formulaes
marwalix
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