How do you factor #0.064a^3+8y^9#?

1 Answer
Apr 12, 2015

When asked to factor #0.064a^3+8y^9#, the first things I notice are the addition and the #a^3# -- a cube.
Then "Oh, look, #8# is a cube and, so is #y^9#".

(#2^3=8# and #(y^3)^3 = y^9#)

So, I'm thinking maybe it's a sum of two cubes.

But what about that #0.064#?

Well, 3 right of the decimal is 1000ths and that's dividing by 1,000 which is #10 ^3#, that is: # color(white)(64)/1000 = color(white)(64)/10^3#

OK, so #0.064 = 64/100 = 64/10^3#

So all I'm left wondering about is the #64#

I cross my fingers and start trying numbers.
#2^3# is only #8#,
#3^3# is an odd number, it can't be #64#
#4^3 = 4*16=64# and there it is:

#0.064a^3+8y^9 = color(red) ((4/10a)^3) + (2y^3)^3#

Use the rule I've memorized:

#color(red)(u^3)+v^3 = ( color(red)(u)+v) (color(red)(u^2) - color(red)(u) v+v^2)# to get:

# color(red)((4/10a)^3) + (2y^3)^3 = [ color(red)((4/10a)) + (2y^3)] [ color(red)((4/10a)^2) + color(red)((4/10a))(2y^3) +(2y^3)^2]#

#= [ (4/10a) + (2y^3)] [ (16/100 a^2) + (4/10a)(2y^3) +(4y^6)]#

#= [ 0.4a + 2y^3] [ 0.16 a^2 + 0.4 a*2y^3 +4y^6]#

#= [ 0.4a + 2y^3] [ 0.16 a^2 + 0.8 ay^3 +4y^6]#.