June 2005 Further Maths

[anillatoo]

Paper 1 Number 4 first part, where I have to show that the sum of the cubes of the roots is -3. I know the sum of the roots and the sum of the squares of the roots but how do I find the sum of the cubes? Any help please? Here is the question...

Show that the sum of the cubes of the roots of the equation
x3 + λ x + 1 = 0
is −3.

 

[Zohaib Sherazi]

If the roots are a,b,c, from the polynomial you have
a+b+c = 0
ab + bc + ca = λ
abc = -1

(a+b+c)^3
= a^3 + b^3 + c^3 + 3a^2(b+c) + 3b^2(c+a) + 3c^2(a+b) + 6abc
a+b+c = 0
so b+c = -a, c+a = -b, a+b = -c

(a+b+c)^3 = -2(a^3 + b^3 + c^3) + 6abc

0 = -2(a^3 + b^3 + c^3) - 6

a^3 + b^3 + c^3 = -3

Alternatively:

Let r1, r2, and r3 be the roots of the polynomial p(x) = 1 + λx + x³. By Viète's formulas, we have

........-r₁r₂ r₃ + (r₁ r₂ + r₁ r₃ + r₂ r₃) x + (-r₁ - r₂ - r₃) x² + x³ = p(x).

Therefore, r₁+ r₂+ r₃= 0, (r₁r₂ + r₁r₃ + r₂r₃) = λ, and r₁r₂r₃= -1. Hence,

........( r₁)³ + ( r₂)³ + ( r₃)³ =

........( r₁+ r₂+ r₃)³ - 3( r₁+ r₂ + r₃)( r₁r₂+ r₁r₃+ r₂r₃) + 3( r₁r₂r₃) =

........0 - 3(0)(λ) + 3(-1) = -3.

 

[anillatoo]

Alright, thanks about that!