Factoring the Sum or Difference of Cubes - Problem 2 3,714 views
Factoring a trinomial. So for this particular problem we are trying to factor 27x³ plus y to the sixth. And just looking at it, trying to figure out what kind of formula we can use.
We know that we have two things, we’re not subtracting so this can’t be the difference of squares, I need to think about what other things we have at our disposal, okay? We do have formulas for the sum of cubes and so what we need to figure out is does this fit the mode?
So remember the sum of cubes is going to be something cubed plus something else cubed. And the trick is figuring out if this fits that formula. So looking at it, 27 is actually 3³, so this becomes 3³ and x³ is obviously x cubed, so this is x³, or what we actually have here is the quantity 3x to the third, okay? So this term 27³ is actually something cubed.
Y to the sixth, remember we're dealing with exponents power to a power we multiply, this is actually going to be the same thing as y² to the third. So what we actually have is 3x³ plus y² cubed and we actually have the sum of cubes.
So where we have to go from here is remembering our sum of cubes formula; b³ plus b³, we don’t need that parenthesis, is going to be equal to a plus b. The first sign agrees, a² minus ab, the second sign is opposite and the last sign is always positive.
So now we just have to plug in 3x wherever we see a and y² whenever we see b. So this then would factor to a, 3x plus b, y² plus a² which is going to be 3x quantity squared, which turns into 9x², minus 3x times y², plus b², so plus this y². Y² times y² will be y to the fourth, so plus y to the fourth. And what we have is 27x³ plus y to the sixth factored down.
So the first thing we had to do was just make sure we can figure out what exactly each of these components is when we cube them to make sure it fits this mode the sum of cubes.