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If a,b,c,d,e, and f are integers for which 1000x^3+27= (ax^2 + bx +c )(d x^2 +ex + f) for all x, then what is a^2+b^2+c^2+d^2+e^2+f^2?

 Aug 30, 2019
 #1
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1000x^3  + 27       can be factored as a sum of cubes thusly :

 

(10x + 3) ( 100x^2  - 30x + 9) 

 

So

 

a  = 0 , b  = 10, c = 3, d = 100, e = -30  and f = 9

 

So

 

(0)^2 + 10^2  + 3^2  + 100^2 + (-30)^2 + 9^2  =  11090

 

CORRECTED ANSWER>>>>Thanks, Alan for catching my error  !!!

 

 

cool cool cool

 Aug 30, 2019
edited by CPhill  Aug 30, 2019

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