IIT JEE , AIEEE Forum - Mathematics , Algebra

konichiwa2x

let

be complex numbers such that

and

Find all possible values of

.

akhil_o

3...
plz dont blame me but i am not able to prove that it is only
1,w,w^2

Conjurer

( a² + b² + c² )/2.( a³ + b³ + c³ )/3= a⁵ + b⁵ + c⁵/5
When a+b+c=0 (proved here http://www.goiit.com/posts/list/algebra-roots-of-polynomial-45973.htm#227419)

=>
(ab+bc+ca)=0
and a+b+c=0 and 3abc=3 =>abc=1

> a,b,c are the roots of the equation:

x^3 -1=0

So we get 1,w+w^2

EDIT: And the answer so is 3.

hsbhatt

Let a,b, and c be the roots of the cubic

x³+px²+qx+r = 0

Then p =-(a+b+c) = 0.

Since

a = 0

a³ = 3abc = 3 (given)

Hence abc =1.
This gives r =-1

Hence the cubic is x³+qx-1 = 0 satisfied by a,b and c.

Hence a³+qa-1 = 0 etc.

Hence a⁵+qa³-a² = 0

Hence

a⁵+q

a³-

a² = 0

Now

a²= -2

ab = -2q

Hence 0+3q+2q = 0

Hence q =0

The eqn is x³-1 = 0, which has the roots 1,

,

²

Hence a²⁰⁰⁷+b²⁰⁰⁷+c²⁰⁰⁷= 1+

+

² = 0

PS: Nice work conjurer. I remembered that but I was half-way thru anyways. Learning slows with age

konichiwa2x

, We have

Now let

Thus

are the roots of

so

So

Therefore

Hence

are the roots of the polynomial

So

Thus since

is divisible by

, we have

Quote from hsbhatt sir:

Hence a²⁰⁰⁷+b²⁰⁰⁷+c²⁰⁰⁷= 1+

+

² = 0

sir, how is it 0?

hsbhatt

my brain was all addled up at that time of the night. So bad that I took 2006 as a multiple of 3. But I thought conjurer has killed the problem effectively isnt it. I mean the solution you posted is not different from mine except for that howler, while conjurer's is damn good.

konichiwa2x

yeah i agree...quite clever. Sometimes we are so stuck upon the "usual" method, we miss smarter stuff.

Ask & Discuss Questions with Community & Experts