quick question (polynomials) (1 Viewer)

[blackfriday]

hey fellas

in arnold ex 4.3 last question, they ask for the value of a^5+b^5+c^5 for x^3+qx+r. now i know how to do it the obvious way but can anyone tell me a really quick way of doing it because i cant be stuffed expanding out to the power of 5.

cheers

 

[Slidey]

x^3=-qx-r
x^5=-qx^3-rx^2
Sub in x= a,b,c and add each result:
a^5+b^5+c^5=-q(a^3+b^3+c^3)-r(a^2+b^2+c^2)

but first:
a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ac)=-2q
and for a^3+b^3+c^3:
x^3=-qx-r
a^3+b^3+c^3=-q(a+b+c)-r=-r

Therefore,
a^5+b^5+c^5=-q(a^3+b^3+c^3)-r(a^2+b^2+c^2)
a^5+b^5+c^5=-q.-r - r.-q=2rq

Answer:
a^5+b^5+c^5 = 2rq #

 

[JamiL]

x^3+qx+r or u can sub a,b n c in...
hence a^3+qa+r = 0 (obviouly =0 cos its a factor/root)
therefore a^3=-qa-r
times by a^2
you get a^5 = -qa^3- ra^2
therefore b^5 = -qb^3- rb^2
therefore c^5 = -qc^3- rc^2
therefore a^5+b^5+c^5 = -q (a^3+b^3+c^3) - r (a^2+b^2+c^2)
the u find a^3+b^3+c^3, and a^2+b^2+c^2 which are easy and if u believe slid rule
we get a^5+b^5+c^5 = 2rq

lol i jus realised u did the exact same thin slide rule... lol
oh well now u have it done 2wice

 

[blackfriday]

um..ah thanks fellas. still a little confused but i'll wrap my skull around it

 

[Slidey]

I made a silly mistake.

a^5+b^5+c^5=-q(a^3+b^3+c^3)-r(a^2+b^2+c^2)
a^5+b^5+c^5=-q.-r - r.-q=2rq

The last line should read:

a^5+b^5+c^5=-2q.-r - r.-q=rq

My apologies.

 

[Trev]

I made a silly mistake.

My apologies.

*Adds to sig.*