# algebraic problem

• Jun 23rd 2010, 03:57 PM
alexandros
algebraic problem
Challenge problem:

Express the following algebraic quantities in terms of the variables λ,μ :

1)$\displaystyle a^2+b^2+c^2$

2)$\displaystyle a^3+b^3+c^3$

3)$\displaystyle a^5+b^5+c^5$

Where $\displaystyle a\neq b\neq c$ are the roots of the following equation:

$\displaystyle x^3+\lambda x+\mu=0$ ,$\displaystyle \lambda\neq 0,\mu\neq 0$ and real Nos

Moderator edit: Approved Challenge question.
• Jun 23rd 2010, 10:20 PM
simplependulum
(a) $\displaystyle a^2 + b^2 + c^2 = (a+b+c)^2 - 2(ab+bc+ca) = -2A$

(b) we have $\displaystyle a^3 + A a + B = 0$

$\displaystyle a^3 = -Aa - B$

$\displaystyle a^3 + b^3 + c^3 = -A(a+b+c) - 3B = -3B$

(c) $\displaystyle a^5 + Aa^3 + Ba^2 = 0$

$\displaystyle a^5 = -Aa^3 - Ba^2$

$\displaystyle a^5 + b^5 + c^5 = -A(a^3 + b^3 + c^3) - B ( a^2 + b^2 + c^2) = 3AB + 2AB = 5AB$
• Jun 24th 2010, 07:06 AM
alexandros
Quote:

Originally Posted by simplependulum
(a) $\displaystyle a^2 + b^2 + c^2 = (a+b+c)^2 - 2(ab+bc+ca) = -2A$

(b) we have $\displaystyle a^3 + A a + B = 0$

$\displaystyle a^3 = -Aa - B$

$\displaystyle a^3 + b^3 + c^3 = -A(a+b+c) - 3B = -3B$

(c) $\displaystyle a^5 + Aa^3 + Ba^2 = 0$

$\displaystyle a^5 = -Aa^3 - Ba^2$

$\displaystyle a^5 + b^5 + c^5 = -A(a^3 + b^3 + c^3) - B ( a^2 + b^2 + c^2) = 3AB + 2AB = 5AB$

O.K correct ,but why don't you show how you get that :

a+b+c=0 and ab+ac+bc = 2A ? Not for me but for other people that want to know the solution of the problem?