Given 7X4−14X3−7X+2=f∈R[X], find the sum of the cubed roots.
Let x1,x2,x3,x4∈R be the roots. Then the polynomial X4−2X3−X+2/7 would have the same roots. If we write the polynomial as X4+a1X3+a2X2+a3X+a4 then per Viete's theorem:
ak=(−1)kσk(x1,x2,x3,x4),k∈{1,2,3,4}, where σk is the k-th elementary symmetrical polynomial. Therefore:
x1+x2+x3+x4=2
x1x2+x1x3+x1x4+x2x3+x2x4+x3x4=0 (∗)
x1x2x3+x1x2x4+x1x3x4+x2x3x4=1
x1x2x3x4=2/7
Now how to determine the sum of the cubed roots?
23=8=(x1+x2+x3+x4)(x1+x2+x3+x4)2=(x1+x2+x3+x4)(x21+x22+x23+x24+2(∗))
Here's where things go out of hand:
(x1+x2+x3+x4)(x21+x22+x23+x24)=(x31+x32+x33+x34)+x21(x2+x3+x4)+x22(x1+x3+x4)+x23(x1+x2+x4)+x24(x1+x2+x3)=8
What should I do here?
Answer
Let
A=x1+x2+x3+x4=2
B=x1x2+x1x3+x1x4+x2x3+x2x4+x3x4=0
C=x1x2x3+x1x2x4+x1x3x4+x2x3x4=1
D=x1x2x3x4=27.
E=x21x2+x1x22+x21x3+x1x23+x21x4+x1x24+x22x3+x2x23+x22x4+x2x24+x23x4+x3x24
We have
A3=x31+x32+x33+x34+3E+6C
and
AB=E+3C.
So,
x31+x32+x33+x34=A3−3(AB−3C)−6C=11.
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