The Three Rooty Polynomial

Algebra Level 4

Let r , s r, s , and t t be the three roots of the equation 8 x 3 + 1001 x + 2008 = 0. 8x^3 + 1001x + 2008 = 0.

Find ( r + s ) 3 + ( s + t ) 3 + ( t + r ) 3 (r + s)^3 + (s + t)^3 + (t + r)^3


The answer is 753.

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Parth Lohomi by Parth Lohomi , 20, India

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3 solutions

Parth Lohomi
Apr 16, 2015

By Vieta's formulas , we have r + s + t = 0 r+s+t = 0 , and so the desired answer is ( r + s ) 3 + ( s + t ) 3 + ( t + r ) 3 = ( 0 − t ) 3 + ( 0 − r ) 3 + ( 0 − s ) 3 = − ( r 3 + s 3 + t 3 ) (r+s)^3+(s+t)^3+(t+r)^3=(0-t)^3+(0-r)^3+(0-s)^3=-(r^3 + s^3 + t^3) . Additionally, using the factorization r 3 + s 3 + t 3 − 3 r s t = ( r + s + t ) ( r 2 + s 2 + t 2 − r s − s t − t r ) = 0 r^3 + s^3 + t^3 - 3rst = (r+s+t)(r^2 + s^2 + t^2 - rs - st - tr) = 0 we have that r 3 + s 3 + t 3 = 3 r s t r^3 + s^3 + t^3 = 3rst . By Vieta's again, r s t = − 2008 8 = − 251 ⟹ − ( r 3 + s 3 + t 3 ) = − 3 r s t = 753 . rst = \dfrac{-2008}8 = -251 \Longrightarrow -(r^3 + s^3 + t^3) = -3rst = \boxed{753}.

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Moderator note:

Nice use of Newton's Identities!

Is it really worth LEVEL 5.What do you think bro...isn't it overrated

Mohit Gupta - 5 years, 6 months ago

Simple use of Vieta’s give us r+s+t=0, r*s*t=-2008/8=-251, gives r+s=-t, s+t=-r, t+r=-s. \text{Simple use of Vieta's give us r+s+t=0, r*s*t=-2008/8=-251, }\\\text{gives r+s=-t, s+t=-r, t+r=-s.} ∴ G i v e n E x p . = − ( t 3 + r 3 + t 3 ) . B u t − ( t 3 + r 3 + t 3 ) + 3 ∗ r ∗ s ∗ t = − ( r + s + t ) ∗ f ( r , s , t ) = 0 ∗ f ( r , s , t ) = 0. ∴ − ( t 3 + r 3 + t 3 ) + 3 ∗ r ∗ s ∗ t = 0 ⟹ E x p . = − ( t 3 + r 3 + t 3 ) = − 3 ∗ r ∗ s ∗ t = − 3 ( − 251 ) = 753 \therefore~Given~Exp.=-(t^3+r^3+t^3). \\But -(t^3+r^3+t^3)+3*r*s*t\\=-(r+s+t)*f(r,s,t)=0*f(r,s,t)=0.\\\therefore~-(t^3+r^3+t^3)+3*r*s*t=0~\\\implies~Exp.=-(t^3+r^3+t^3)=-3*r*s*t=-3(-251)=\color{#D61F06}{\boxed{753}}

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Rick B
Apr 22, 2015

By Vieta's, r + s + t = 0 r+s+t = 0 , so we can rewrite ( r + s ) 3 + ( s + t ) 3 + ( t + r ) 3 (r+s)^3+(s+t)^3+(t+r)^3

as ( 0 − t ) 3 + ( 0 − r ) 3 + ( 0 − s ) 3 = − t 3 − r 3 − s 3 = − ( t 3 + r 3 + s 3 ) (0-t)^3+(0-r)^3+(0-s)^3 = -t^3-r^3-s^3 = -(t^3+r^3+s^3)

Let P k = r k + s k + t k P_k = r^k+s^k+t^k

By Newton's Sums, 8 P 3 + 3 ⋅ 2008 = 0 ⟹ P 3 = − 3 ⋅ 2008 8 = − 753 8P_3 + 3 \cdot 2008 = 0 \implies P_3 = -\dfrac{3 \cdot 2008}{8} = -753

So the answer is − P 3 = 753 -P_3 = \boxed{753}

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