Factoring

What would  x^6-y^6 be when completely factored?

$$\textcolor[rgb]{1,0,0}{ a^3 + b^3 = (a + b)(a^2-ab + b^2) }\\ \textcolor[rgb]{1,0,0}{ a^3 - b^3 = (a - b)(a^2 + ab + b^2) } \\ (a^6-b^6)=(a^3 + b^3)*(a^3 - b^3 )=(a + b)(a^2-ab + b^2)(a - b)(a^2 + ab + b^2)$$

$$a=x \text{ and } b=y$$ $$(x^6-y^6)=(x + y)(x^2-xy + y^2)(x - y)(x^2 + xy + y^2)$$

There are a few factoring forms worth knowing, three of which are:

--- Difference of squares:  A² - B²  =  (A + B)(A - B)

--- Difference of cubes:     A³ - B³  =  (A - B)(A² + AB + B²)

--- Sum of cubes:             A³ + B³  =  (A + B)(A² - AB + B²)

In  x^6 - y^6,  first use difference of squares:

  =  (x³)² - (y³)²  to get:  (x³ + y³)(x³ - y³)    

Now use both sum of cubes and difference of cubes:

  =  [ (x - y)(x² + xy + y²) ][ (x + y)(x² - xy + y²) ]

  =  (x - y)(x + y)(x² + xy + y²)(x² - xy + y²)

Thank you!