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?
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
(0)^2 + 10^2 + 3^2 + 100^2 + (-30)^2 + 9^2 = 11090
CORRECTED ANSWER>>>>Thanks, Alan for catching my error !!!