SOLUTION: Factor both equations (difference and sum of cubes) 1. (x-3)^3 + (3x-2)^3 2. (x^9/512) - 512

Algebra ->  Equations -> SOLUTION: Factor both equations (difference and sum of cubes) 1. (x-3)^3 + (3x-2)^3 2. (x^9/512) - 512      Log On


   

Question 1165761: Factor both equations (difference and sum of cubes)
1. (x-3)^3 + (3x-2)^3
2. (x^9/512) - 512
Found 3 solutions by greenestamps, Boreal, MathTherapy:
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


You apparently know the patterns -- use them.

Note %28x%5E3%29%5E3+=+x%5E9; 8%5E3+=+512

Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
a^3+b^3 factors into (a+b) (a^2-ab+b^2)
so {(x-3)+(3x-2)}* {(x-3)^2)-(x-3)(3x-2)+(3x-2)^2}
the other is
{(x3/8)-8} * ((x^6/64)+x^3+64)}
but the first can be factored as well
{(x/2)-2} * {(x^2/4)+x+4}
and the whole thing is
{(x/2)-2} * {(x^2/4)+x+4}*{((x^6/64)+x^3+64)}

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
Factor both equations (difference and sum of cubes)
1. (x-3)^3 + (3x-2)^3
2. (x^9/512) - 512
1.  

2.