Factorization is an art

Algebra Level 5

A root of 10 x 3 − 3 x 2 − 3 x − 1 = 0 10x^3-3x^2-3x-1=0 is of the form a 3 + b 3 + 1 c \dfrac{\sqrt[3]{a}+\sqrt[3]{b}+1}{c} , where a a , b b , and c c are positive integers. Find a + b + c a+b+c .


The answer is 142.

Tommy Li by Tommy Li , 22, Hong Kong

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2 solutions

Tommy Li
Jun 5, 2016

10 x 3 − 3 x 2 − 3 x − 1 = 0 10x^3-3x^2-3x-1=0

11 x 3 = x 3 + 3 x 2 + 3 x + 1 11x^3=x^3+3x^2+3x+1

11 x 3 = ( x + 1 ) 3 11x^3=(x+1)^3

11 = ( x + 1 ) 3 x 3 11=\frac{(x+1)^3}{x^3}

11 3 = 1 + 1 x \sqrt[3]{11}=1+\frac{1}{x}

1 11 3 − 1 = x \frac{1}{\sqrt[3]{11}-1}=x

1 ( ( 11 3 ) 2 + 11 3 + 1 ) ( 11 3 − 1 ) ( ( 11 3 ) 2 + 11 3 + 1 ) = x \frac{1((\sqrt[3]{11})^2+\sqrt[3]{11}+1)}{(\sqrt[3]{11}-1)((\sqrt[3]{11})^2+\sqrt[3]{11}+1)}=x

121 3 + 11 3 + 1 11 − 1 = x \frac{\sqrt[3]{121}+\sqrt[3]{11}+1}{11-1}=x

121 3 + 11 3 + 1 10 = x \frac{\sqrt[3]{121}+\sqrt[3]{11}+1}{10}=x

a + b + c = 121 + 11 + 10 = 142 a+b+c=121+11+10=142

19 Helpful 1 Interesting 2 Brilliant 0 Confused

Same Method (+1) :)

Aditya Sky - 5 years ago

− ( X 3 − 3 X 2 − 3 X − 1 ) = − ( X + 1 ) 3 . ∴ 10 X 3 − 3 X 2 − 3 X − 1 = 11 X 3 − ( X + 1 ) 3 = 0. ⟹ ( 1 3 1 ∗ X ) 3 − ( X + 1 ) 3 = 0. D i f f e r e n c e o f t w o c u b e s . ∴ ( A ) { 11 3 ∗ X − X − 1 } = 0 , o r ( B ) { ( 1 3 1 ∗ X ) 2 + 1 3 1 ∗ X ∗ ( X + 1 ) + ( X + 1 ) 3 } = 0. ( A ) ⟹ X ( 11 3 − 1 ) = 1. ∴ X = 1 11 3 − 1 . B u t g e t t i n g r i d o f t h e r a d i c a l i n t h e d e n o m i n a t o r , X = ( 11 3 ) 2 + 11 3 ∗ 1 + 1 2 ( 11 3 ) 3 − 1 3 ) = 121 3 + 11 3 + 1 11 − 1 = a 3 + b 3 + 1 c . ∴ a + b + c = 121 + 11 + 10 = 142. -(X^3-3X^2-3X-1)=-(X+1)^3.\\ \therefore~10X^3-3X^2-3X-1=11X^3-(X+1)^3=0.\\ \implies~(\sqrt[3]11*X)^3-(X+1)^3=0.~~~~~Difference~ of~ two~ cubes.\\ \therefore~~(A)~~\{\sqrt[3]{11}*X-X-1\}=0,~~~or~~~~~(B)~~\{(\sqrt[3]11*X)^2+\sqrt[3]11*X*(X+1)+(X+1)^3\}=0.\\ (A) ~~~\implies~ X( \sqrt[3]{11}-1)=1.\\ \therefore~X=\dfrac 1{ \sqrt[3]{11}-1}.~~But~getting~ rid~ of~ the~ radical~ in ~the~ denominator, \\ X=\dfrac{ ( \sqrt[3]{11})^2 + \sqrt[3]{11}*1+1^2}{( \sqrt[3]{11})^3-1^3)}\\ = \dfrac{ \sqrt[3]{121} + \sqrt[3]{11}+1}{11-1}\\ = \dfrac{ \sqrt[3] a + \sqrt[3] b+1} c. \\ \therefore~~a+b+c=121+11+10=\Large ~~~\color{#D61F06}{142}.

2 Helpful 1 Interesting 0 Brilliant 0 Confused

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