ABC's are hauntin' you

Algebra Level 4

If a + b + c = 0 , \Large \color{#D61F06}{a}+\color{#3D99F6}{b}+\color{#20A900}{c} = 0, then a 5 + b 5 + c 5 a b + b c + c a = K ⋅ a b c . \ \Large \dfrac{\color{#D61F06}{a}^5 + \color{#3D99F6}{b}^5 + \color{#20A900}{c} ^5}{\color{#D61F06}{a}\color{#3D99F6}{b} + \color{#3D99F6}{b}\color{#20A900}{c} +\color{#20A900}{c} \color{#D61F06}{a}}=K\cdot \color{#D61F06}{a}\color{#3D99F6}{b}\color{#20A900}{c}. K = ? \Large K \ = \ ?

Note: Assume that a , b , c a,b,c are not all 0.


The answer is -5.

Lakshya Singh by Lakshya Singh , 19, India

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3 solutions

Ricardo Takayama
Nov 24, 2015

Using Newton's Identities for a cubic polynomial:

P 0 = a 0 + b 0 + c 0 = 3 P 1 = a 1 + b 1 + c 1 = 0 P 2 = a 2 + b 2 + c 2 = ( a + b + c ) ( a + b + c ) − 2 ( a b + b c + c a ) = − 2 ( a b + b c + c a ) P 3 = a 3 + b 3 + c 3 = ( a + b + c ) ( a 2 + b 2 + c 2 ) − ( a b + b c + c a ) ( a + b + c ) + 3 a b c = 3 a b c P 4 = a 4 + b 4 + c 4 = ( a + b + c ) ( a 3 + b 3 + c 3 ) − ( a b + b c + c a ) ( a 2 + b 2 + c 2 ) + a b c ( a + b + c ) P 5 = a 5 + b 5 + c 5 = ( a + b + c ) ( a 4 + b 4 + c 4 ) − ( a b + b c + c a ) ( a 3 + b 3 + c 3 ) + a b c ( a 2 + b 2 + c 2 ) = − ( a b + b c + c a ) ( 3 a b c ) + a b c ( − 2 ( a b + b c + c a ) ) = − 5 a b c ( a b + b c + c a ) \begin{array} { l l l} P_0 & = a ^ 0 + b ^ 0 + c ^ 0 & = 3 \\ P_1 & = a ^ 1 + b ^ 1 + c ^ 1 & = 0 \\ P_2 & = a ^ 2 + b ^ 2 + c ^ 2 & = (a + b + c)(a + b + c) - 2 ( a b + b c + c a)\\ & & = - 2 ( a b + b c + c a) \\ P_3 & = a ^ 3 + b ^ 3 + c ^ 3 & = (a + b + c) (a^2 + b^2 + c^2) - ( a b + b c + c a)(a + b + c) + 3 a b c \\ & & = 3 a b c \\ P_4 & = a ^ 4 + b ^4 + c ^ 4 & = (a + b + c) (a^3 + b^3 + c^3) - ( a b + b c + c a)(a ^2 + b ^2 + c ^2) + a b c ( a + b + c) \\ P_5 & = a ^ 5 + b ^5 + c ^ 5 & = (a + b + c) (a^4 + b^4 + c^4) - ( a b + b c + c a)(a ^3 + b ^3 + c ^3) + a b c ( a ^ 2 + b ^ 2 + c ^ 2) \\ & & = - ( a b + b c + c a) (3 a b c) + a b c (- 2 ( a b + b c + c a)) = - 5 a b c ( a b + b c + c a)\\ \end{array}

Hence, a 5 + b 5 + c 5 a b + b c + c a = − 5 a b c ; K = − 5 \dfrac{a ^ 5 + b ^5 + c ^ 5}{ a b + b c + c a} = - 5 a b c ; K = -5

3 Helpful 0 Interesting 1 Brilliant 0 Confused
Shreyash Rai
Nov 30, 2015

will this be fine?

2 Helpful 0 Interesting 1 Brilliant 0 Confused

Yes, it is :)

Ash Ketchup - 4 months, 1 week ago
Lakshya Singh
Dec 1, 2015

A very nice solution

0 Helpful 0 Interesting 0 Brilliant 0 Confused

And a cool problem

Shreyash Rai - 5 years, 6 months ago

Hey shreyash can you post some mind boggling questions in algebra?

Lakshya Singh - 5 years, 6 months ago

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I can try but im not much of a maker just a solver

Shreyash Rai - 5 years, 6 months ago

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