Show that the sum of the cubes of the roots of the equation

x3 + λ x + 1 = 0

is −3.

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- Thread starter anillatoo
- Start date

Show that the sum of the cubes of the roots of the equation

x3 + λ x + 1 = 0

is −3.

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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.