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Finding values a, b, c, d, e, f

mente oscura

Well-known member
Nov 29, 2013
172
Hello.

Finding values: a, b, c, d, e, f; for:


[tex]a+b+c+d+e+f=-6[/tex]

[tex]a^2+b^2+c^2+d^2+e^2+f^2=534[/tex]

[tex]a^3+b^3+c^3+d^3+e^3+f^3=-4662[/tex]

[tex]a^4+b^4+c^4+d^4+e^4+f^4=158886[/tex]

[tex]a^5+b^5+c^5+d^5+e^5+f^5=-2104806[/tex]

[tex]a^6+b^6+c^6+d^6+e^6+f^6=51872694[/tex]


Regards.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,708
For $n=1,2,3,4,5,6$, write $P_n = a^n+b^n+c^n+d^n+e^n+f^n$. Also, let $E_n$ be the $n$th elementary symmetric polynomial in $a,b,c,d,e,f.$ We are told that $$P_1 = -6,\quad P_2 = 534,\quad P_3 = -4662,\quad P_4 = 158886,\quad P_5 = -2104806,\quad P_6 = 51872694.$$ It follows from Newton's identities that $$\begin{aligned}\phantom{1}E_1 &= P_1 = -6, \\ 2E_2 &= E_1P_1 - P_2 \\ &= 36 - 534 = -498, \text{ so }E_2 = -249, \\ 3E_3 &= E_2P_1 - E_1P_2 + P_3 \\ &= 1494 + 3204 - 4662 = 36, \text{ so }E_3 = 12, \\ 4E_4 &= E_3P_1 - E_2P_2 + E_1P_3 - P_4 \\ &= -72 + 132966 + 27972 - 158886 = 1980, \text{ so }E_4 = 495, \\ 5E_5 &= E_4P_1 - E_3P_2 + E_2P_3 - E_1P_4 + P_5 \\ &= -2970 - 6408 + 1160838 + 953886 - 2104806 = 30, \text{ so }E_5 = -6, \\ 6E_6 &= E_5P_1 - E_4P_2 + E_3P_3 - E_2P_4 + E_1P_5 - P_6 \\ &= 36 - 264330 - 55944 + 39562614 + 12628836 - 51872694 = -1482, \text{ so }E_6 = -247. \end{aligned}$$ Then the equation with roots $a,b,c,d,e,f$ is $x^6 - E_1x^5 + E_2x^4 - E_3x^3 + E_4x^2 - E_5x + E_6 = 0,$ or $x^6 + 6x^5 - 249x^4 - 12x^3 + 495x^2 + 6x - 247 = 0.$ Since $247 = 13\times 19,$ it is not hard to factorise that as $(x^2-1)^2(x-13)(x+19) = 0.$ Thus the numbers $a,b,c,d,e,f$ are (in some order) $1,1,-1,-1,13,-19.$
 
Last edited:

mente oscura

Well-known member
Nov 29, 2013
172
For $n=1,2,3,4,5,6$, write $P_n = a^n+b^n+c^n+d^n+e^n+f^n$. Also, let $E_n$ be the $n$th elementary symmetric polynomial in $a,b,c,d,e,f.$ We are told that $$P_1 = -6,\quad P_2 = 534,\quad P_3 = -4662,\quad P_4 = 158886,\quad P_5 = -2104806,\quad P_6 = 51872694.$$ It follows from Newton's identities that $$\begin{aligned}\phantom{1}E_1 &= P_1 = -6, \\ 2E_2 &= E_1P_1 - P_2 \\ &= 36 - 534 = -498, \text{ so }E_2 = -249, \\ 3E_3 &= E_2P_1 - E_1P_2 + P_3 \\ &= 1494 + 3204 - 4662 = 36, \text{ so }E_3 = 12, \\ 4E_4 &= E_3P_1 - E_2P_2 + E_1P_3 - P_4 \\ &= -72 + 132966 + 27972 - 158886 = 1980, \text{ so }E_4 = 495, \\ 5E_5 &= E_4P_1 - E_3P_2 + E_2P_3 - E_1P_4 + P_5 \\ &= -2970 - 6408 + 1160838 + 953886 - 2104806 = 30, \text{ so }E_5 = -6, \\ 6E_6 &= E_5P_1 - E_4P_2 + E_3P_3 - E_2P_4 + E_1P_5 - P_6 \\ &= 36 - 264330 - 55944 + 39562614 + 12628836 - 51872694 = -1482, \text{ so }E_6 = -247. \end{aligned}$$ Then the equation with roots $a,b,c,d,e,f$ is $x^6 - E_1x^5 + E_2x^4 - E_3x^3 + E_4x^2 - E_5x + E_6 = 0,$ or $x^6 + 6x^5 - 249x^4 - 12x^3 + 495x^2 + 6x - 247 = 0.$ Since $247 = 13\times 19,$ it is not hard to factorise that as $(x^2-1)^2(x-13)(x+19) = 0.$ Thus the numbers $a,b,c,d,e,f$ are (in some order) $1,1,-1,-1,13,-19.$
Hello.

Very well, Opalg, is correct.:)

Regards.