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Problem Of The Week #375 Jul 16th, 2019

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,894
Here is this week's POTW:

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Let $a,\,b$ and $c$ be distinct real numbers such that

$a^3=3(b^2+c^2)-25\\b^3=3(c^2+a^2)-25\\c^3=3(a^2+b^2)-25$

Evaluate $abc$.


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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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anemone

MHB POTW Director
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Feb 14, 2012
3,894
Congratulations to Opalg for his correct solution(Cool), which you can find below:

$$(1)\qquad a^3 = 3(b^2+c^2) - 25$$
$$(2)\qquad b^3 = 3(c^2+a^2) - 25$$
$$(3)\qquad c^3 = 3(a^2+b^2) - 25$$
Subtract (2) from (1): $a^3 - b^3 = 3(b^2 - a^2)$. Since $a$, $b$, $c$ are distinct, we can divide that by $a-b$ to get $$(4)\qquad a^2 + ab + b^2 = -3(a+b).$$ From the cyclic symmetry of equations (1)-(3), it will follow in the same way that $$(5)\qquad b^2 + bc + c^2 = -3(b+c),$$ $$(6)\qquad c^2 + ca + a^2 = -3(c+a).$$ Subtract (5) from (4) to get $a^2 - c^2 + b(a-c) = -3(a-c).$ This time, $a-c$ is a factor, and since it is not zero we can divide by it, getting $a+b+c = -3.$

Using the notation of Newton's identities, let $p_n = a^n+b^n+c^n$ ($n=1,2,3$), $e_1 = p_1 = a+b+c = -3$, $e_2 = bc+ca+ab$ and $e_3 = abc$. Newton's identities state that $$(7)\qquad p_2 = e_1p_1 - 2e_2 = 9 - 2e_2,$$ $$(8)\qquad p_3 = e_1p_2 - e_2p_1 + 3e_3.$$ Also, adding equations (1), (2), (3) gives $$(9)\qquad p_3 = 6p_2 - 75,$$ and adding (4), (5), (6) gives $$(10)\qquad 2p_2+e_2 = -6p_1 = 18.$$ Now compare (7) with (10) to see that $p_2 = 9$ and $e_2 = 0.$ (9) then becomes $p_3 = -21$. Substituting that into (8) then gives $-21 = -27 + 3e_3$. Therefore $\boxed{abc = e_3 = 2}.$

The equation with roots $a,b,c$ is $x^3 - e_1x^2 + e_2x - e_3 = 0$, or $x^3 + 3x^2 - 2 = 0$. That factorises as $(x+1)(x^2 + 2x - 2) = 0$, with roots $-1$, $-1 \pm\sqrt3$. Those are therefore the values of $a,b,c$ (in some order).
 
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