# Video: Factorisation by Grouping Involving Factorising the Difference of Two Cubes and the Difference of Two Squares

Completely factor 𝑥²𝑦³ − 27𝑥² − 𝑦³ + 27.

01:57

### Video Transcript

Completely factor 𝑥 squared 𝑦 cubed minus 27𝑥 squared minus 𝑦 cubed plus 27.

Here we have four terms, so we can factor by grouping. So in order to factor by grouping, we must group the first two terms together and the last two terms together. And now we take a greatest common factor out of each set of parentheses.

Out of the first set, we can take out an 𝑥 squared. And once we take that out, we’re left with 𝑦 cubed minus 27. Out of the next set, we can take out a negative one. And once we do that, we’re left with 𝑦 cubed minus 27. So the parentheses match. That’s good; that’s what we want.

So the GCFs come together to be one factor. And the matching parentheses become the other factor. However, we aren’t completely done because each of these factors simplify more. The first factor of 𝑥 squared minus one is something called a difference of squares. So we if have 𝑎 squared minus 𝑏 squared, it would factor to be 𝑎 plus 𝑏 𝑎 minus 𝑏.

So 𝑥 squared minus one, we would essentially have to square-root each of them to find the 𝑎 and the 𝑏. So the square root of 𝑥 squared is 𝑥, and the square root of one is one. So it would turn into 𝑥 plus one 𝑥 minus one.

Now for the 𝑦 cubed minus 27, this is called a difference of cubes. So if we have 𝑎 cubed minus 𝑏 cubed, it factors to be 𝑎 minus 𝑏 times 𝑎 squared plus 𝑎𝑏 plus 𝑏 squared. So in order to find 𝑎 and 𝑏, we would have to cube-root 𝑎 cubed and cube-root 𝑏 cubed. So if we have 𝑦 cubed minus 27, we need to cube-root 𝑦 cubed and we need to cube-root 27. The cube root of 𝑦 cubed is 𝑦, and the cube root of 27 is three. So we would have 𝑦 minus three times 𝑦 squared plus three 𝑦 plus nine. So now it is completely factor. So our final answer is 𝑥 plus one times 𝑥 minus one times 𝑦 minus three times 𝑦 squared plus three 𝑦 plus nine.