```Question 277514
{{{x^6 - y^6}}}
There are a variety of techniques you learn when factoring. One of these techniques is to factor by using certain patterns:<ul><li>Difference of squares: {{{(a^2 - b^2) = (a+b)(a-b)}}}</li><li>Sum of cubes: {{{(a^3 + b^3) = (a+b)(a^2-ab+b^2)}}}</li><li>Difference of cubes: {{{(a^3 - b^3) = (a-b)(a^2+ab+b^2)}}}</li><li>Perfect square trinomials:<ul><li>{{{a^2 +2ab + b^2 = (a+b)^2}}}</li><li>{{{a^2 -2ab + b^2 = (a-b)^2}}}</li></ul></li></ul>
This problem happens to be one where these patterns are all you need to factor it. Your expression has only two terms. This is too many terms for the last two patterns. And you expression has a "-" between the two terms so we can't use the second pattern (yet).<br>
So the question is, is {{{x^6 - y^6}}} a difference of squares or a difference of cubes? Since exponent that is divisible by 2 is a perfect square and since any exponent that is divisible by 3 is a perfect square, and since 6 divisible by both 2 and 3, the answer to the question is both! So we can start with either pattern. I'll do it both ways.<br>
Difference of squares first:
{{{x^6 - y^6 = (x^3)^2 - (y^3)^2}}}
Using the pattern we get:
{{{(x^3 + y^3)(x^3 - y^3)}}}
When factoring you keep going until you cannot factor any further. The two factors we have now are clearly a sum of cubes and a difference of cubes, respectively. We can use those patterns on them:
{{{green((x+y)(x^2-xy+y^2)(x-y)(x^2 +xy + y^2))}}}<br>
Using difference of cubes first:
{{{x^6 - y^6 = (x^2)^3 - (y^2)^3}}}
Using the pattern we get:
{{{(x^2 - y^2)((x^2)^2 + (x^2)(y^2) + (y^2)^2)}}}
The last factor can be simplified and the first factor is a difference of squares we can factor:
{{{red((x+y)(x- y)(x^4 + x^2y^2 + y^4))}}}
The first two factors match two of the factors we found earlier. Earlier, however, we ended up with four factors and here we only have three. So somehow
the third factor must factor in the other two factors we found earlier: {{{x^2 +xy + y^2}}} and {{{x^2 -xy + y^2}}}. But the factoring involved in turning {{{x^4 + x^2y^2 + y^4)}}} into {{{(x^2 +xy + y^2)(x^2 -xy + y^2)}}} is very complex and I will not do it here. I'll just say then, when faced with choosing between difference of squares and difference of cubes, choose difference of squares first. It makes things easier.
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In response to the note you included in your thank you...
The answer you mention is the same as the one above where we started by using the difference of cubes first. (I changed it so it shows up in red so you can find it more easily.)<br>
But it remains true that this factored expression is not as fully factored as the one you get when you start with the difference of squares. (I changed this one to show up in green.) The red is correct. The green is correct and a better answer.```