Kay G. answered 03/24/14
Tutor
4.9
(34)
~20 Years Accounting Tutoring Experience
As to your third one, that's the difference of 2 cubes. You'll want to start recognizing some cubes, i.e. 8 is 23, 27 is 33, 64 is 43... probably enough for now. (And don't forget that 1 is also 13.)
a3 - b3 = (a - b)(a2 + ab + b2)
Part of the trick here is working with both the numbers and the variables.
It will be easier to make yours 8x3 - 27 so as not to mix up a and the variable. "a" is actually 8x, not just the variable x. That's why I changed it to x. So the "a" consists of both the 8 and the x, both coefficient and variable.
I would start by getting your cubes of everything: 2, x and 3. Every time you have a, that's 2x. Every time you have b, that's 3.
Now let's deal with a2 and b2. If you factor out 2x from 8x3, what do you have left? The squares, hence a2. If you factor out 3 from 27, what do you have left? The square of it, hence b2.
Just see what you can do with that. I hope you have done easier ones than this first, like x3 - 27, since this one is a bit more difficult. It might be easier to do that first, and then go back and deal with the 8 separately, remembering that's part of a and would go wherever a is.
a3 - b3 = (a - b)(a2 + ab + b2)
Part of the trick here is working with both the numbers and the variables.
It will be easier to make yours 8x3 - 27 so as not to mix up a and the variable. "a" is actually 8x, not just the variable x. That's why I changed it to x. So the "a" consists of both the 8 and the x, both coefficient and variable.
I would start by getting your cubes of everything: 2, x and 3. Every time you have a, that's 2x. Every time you have b, that's 3.
Now let's deal with a2 and b2. If you factor out 2x from 8x3, what do you have left? The squares, hence a2. If you factor out 3 from 27, what do you have left? The square of it, hence b2.
Just see what you can do with that. I hope you have done easier ones than this first, like x3 - 27, since this one is a bit more difficult. It might be easier to do that first, and then go back and deal with the 8 separately, remembering that's part of a and would go wherever a is.