Its time for cubes

Algebra Level 5

Let a 1 , a 2 , … , a 4 a_1, a_2, \ldots,a_{4} denote the roots to the equation x 1 + 2 x 2 + 3 x 3 + 4 x 4 = 0 x^1 + 2x^2 + 3x^3 + 4x^4 = 0 . Then,

( 8 a 1 3 + 1 ) ( 8 a 2 3 + 1 ) ⋯ ( 8 a 4 3 + 1 ) = − A B (8a_{ 1 }^{ 3 }+1)(8a_{ 2 }^{ 3 }+1)\cdots (8a_{ 4 }^{ 3 }+1)=-\frac { A }{ B }

where A , B A,B are positive coprime integers. What is A + B A+B ?

Inspiration

You can refer to this


The answer is 51.

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Joel Yip by Joel Yip , 21, Singapore

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2 solutions

Joel Yip
Mar 14, 2016

( 8 a 1 3 + 1 ) ( 8 a 2 3 + 1 ) ( 8 a 3 3 + 1 ) ( 8 a 4 3 + 1 ) = 8 4 ( a 1 3 + 1 8 ) ( a 2 3 + 1 8 ) ( a 3 3 + 1 8 ) ( a 4 3 + 1 8 ) (8a_{ 1 }^{ 3 }+1)(8a_{ 2 }^{ 3 }+1)(8a_{ 3 }^{ 3 }+1)(8a_{ 4 }^{ 3 }+1)={ 8 }^{ 4 }\left( a_{ 1 }^{ 3 }+\frac { 1 }{ 8 } \right) \left( a_{ 2 }^{ 3 }+\frac { 1 }{ 8 } \right) \left( a_{ 3 }^{ 3 }+\frac { 1 }{ 8 } \right) \left( a_{ 4 }^{ 3 }+\frac { 1 }{ 8 } \right)

With some expansion,

r 3 + 1 8 = ( r + 1 2 ) ( r − 1 + 3 i 4 ) ( r − 1 − 3 i 4 ) { r }^{ 3 }+\frac { 1 }{ 8 } =\left( r+\frac { 1 }{ 2 } \right) \left( r-\frac { 1+\sqrt { 3 } i }{ 4 } \right) \left( r-\frac { 1-\sqrt { 3 } i }{ 4 } \right)

8 4 ( a 1 + 1 2 ) ( a 1 − 1 + 3 i 4 ) ( a 1 − 1 − 3 i 4 ) ( a 2 + 1 2 ) ( a 2 − 1 + 3 i 4 ) ( a 2 − 1 − 3 i 4 ) ( a 3 + 1 2 ) ( a 3 − 1 + 3 i 4 ) ( a 3 − 1 − 3 i 4 ) ( a 4 + 1 2 ) ( a 4 − 1 + 3 i 4 ) ( a 4 − 1 − 3 i 4 ) { 8 }^{ 4 }\left( { a }_{ 1 }+\frac { 1 }{ 2 } \right) \left( { a }_{ 1 }-\frac { 1+\sqrt { 3 } i }{ 4 } \right) \left( { a }_{ 1 }-\frac { 1-\sqrt { 3 } i }{ 4 } \right) \left( { a }_{ 2 }+\frac { 1 }{ 2 } \right) \left( { a }_{ 2 }-\frac { 1+\sqrt { 3 } i }{ 4 } \right) \left( { a }_{ 2 }-\frac { 1-\sqrt { 3 } i }{ 4 } \right) \left( { a }_{ 3 }+\frac { 1 }{ 2 } \right) \left( { a }_{ 3 }-\frac { 1+\sqrt { 3 } i }{ 4 } \right) \left( { a }_{ 3 }-\frac { 1-\sqrt { 3 } i }{ 4 } \right) \left( { a }_{ 4 }+\frac { 1 }{ 2 } \right) \left( { a }_{ 4 }-\frac { 1+\sqrt { 3 } i }{ 4 } \right) \left( { a }_{ 4 }-\frac { 1-\sqrt { 3 } i }{ 4 } \right)

In This note , I generalised that type of function and let f ( x ) = ∑ m = 1 4 m x m f\left( x \right) =\sum _{ m=1 }^{ 4 }{ m{ x }^{ m } }

We rearrange and 8 4 ( a 1 + 1 2 ) ( a 2 + 1 2 ) ( a 3 + 1 2 ) ( a 4 + 1 2 ) ( a 1 − 1 + 3 i 4 ) ( a 2 − 1 + 3 i 4 ) ( a 3 − 1 + 3 i 4 ) ( a 4 − 1 + 3 i 4 ) ( a 1 − 1 − 3 i 4 ) ( a 2 − 1 − 3 i 4 ) ( a 3 − 1 − 3 i 4 ) ( a 4 − 1 − 3 i 4 ) { 8 }^{ 4 }\left( { a }_{ 1 }+\frac { 1 }{ 2 } \right) \left( { a }_{ 2 }+\frac { 1 }{ 2 } \right) \left( { a }_{ 3 }+\frac { 1 }{ 2 } \right) \left( { a }_{ 4 }+\frac { 1 }{ 2 } \right) \left( { a }_{ 1 }-\frac { 1+\sqrt { 3 } i }{ 4 } \right) \left( { a }_{ 2 }-\frac { 1+\sqrt { 3 } i }{ 4 } \right) \left( { a }_{ 3 }-\frac { 1+\sqrt { 3 } i }{ 4 } \right) \left( { a }_{ 4 }-\frac { 1+\sqrt { 3 } i }{ 4 } \right) \left( { a }_{ 1 }-\frac { 1-\sqrt { 3 } i }{ 4 } \right) \left( { a }_{ 2 }-\frac { 1-\sqrt { 3 } i }{ 4 } \right) \left( { a }_{ 3 }-\frac { 1-\sqrt { 3 } i }{ 4 } \right) \left( { a }_{ 4 }-\frac { 1-\sqrt { 3 } i }{ 4 } \right)

8 4 ( − 1 ) 4 f ( − 1 2 ) 4 ( − 1 ) 4 f ( 1 + 3 i 4 ) 4 ( − 1 ) 4 f ( 1 − 3 i 4 ) 4 { 8 }^{ 4 }\frac { { \left( -1 \right) }^{ 4 }f\left( -\frac { 1 }{ 2 } \right) }{ 4 } \frac { { \left( -1 \right) }^{ 4 }f\left( \frac { 1+\sqrt { 3 } i }{ 4 } \right) }{ 4 } \frac { { \left( -1 \right) }^{ 4 }f\left( \frac { 1-\sqrt { 3 } i }{ 4 } \right) }{ 4 }

8 4 ( − 1 8 ) 4 ( − 1 2 + 3 3 8 i ) 4 ( − 1 2 + 3 3 8 i ) 4 = 8 4 ( − 1 8 ) ( 1 4 + 27 64 ) 64 = 8 4 ( − 1 8 ) ( 43 64 ) 64 = − 8 4 43 8 × 64 × 64 = − 43 8 = − 43 2 3 { 8 }^{ 4 }\frac { \left( -\frac { 1 }{ 8 } \right) }{ 4 } \frac { \left( -\frac { 1 }{ 2 } +\frac { 3\sqrt { 3 } }{ 8 } i \right) }{ 4 } \frac { \left( -\frac { 1 }{ 2 } +\frac { 3\sqrt { 3 } }{ 8 } i \right) }{ 4 } ={ 8 }^{ 4 }\frac { \left( -\frac { 1 }{ 8 } \right) \left( \frac { 1 }{ 4 } +\frac { 27 }{ 64 } \right) }{ 64 } \\ ={ 8 }^{ 4 }\frac { \left( -\frac { 1 }{ 8 } \right) \left( \frac { 43 }{ 64 } \right) }{ 64 } \\ =-{ 8 }^{ 4 }\frac { 43 }{ 8\times 64\times 64 } \\ =-\frac { 43 }{ 8 } \\ =-\frac { 43 }{ { 2 }^{ 3 } }

Thus A = 43 A=43 and B = 8 B=8 . So A + B = 43 + 8 = 51 A + B = 43 + 8 = 51 .

7 Helpful 0 Interesting 0 Brilliant 0 Confused
汶良 林
Mar 18, 2016

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Great solution! +1!

Joel Yip - 5 years, 2 months ago

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This is great!

Joel Yip - 5 years, 2 months ago

I don't quite understand how you got the third step?

Is it cubing?

Joel Yip - 5 years, 2 months ago

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Yes. I cube both sides.

汶良 林 - 5 years, 2 months ago

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Thank you! otherwise again great solution!

Joel Yip - 5 years, 2 months ago

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@Joel Yip – can this question be done by taking m= 8a1^3 +1 putting values of a in terms of m and then solving .. i tried this but i am not getting the solution please provide i solution in a way i have described...

Deepansh Jindal - 5 years, 1 month ago

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