G gemini77 New member Joined Mar 5, 2010 Messages 1 Mar 5, 2010 #1 factoring x^6 - y^6 difference of squares is (x-y)(x+y) X^3-y^3 is (x-y)(x^2 + xy + y^2) to ^4 is (x-y)(x+y) (x^2 + xy + y^2) what is to the ^5 or ^6 look like??????
factoring x^6 - y^6 difference of squares is (x-y)(x+y) X^3-y^3 is (x-y)(x^2 + xy + y^2) to ^4 is (x-y)(x+y) (x^2 + xy + y^2) what is to the ^5 or ^6 look like??????
T tutor_joel Junior Member Joined Feb 6, 2010 Messages 157 Mar 5, 2010 #2 you have the right idea. \(\displaystyle x^6-y^6 = (x^3)^2-(y^3)^2\) there's your difference of two squares. Then you have difference and sum of two cubes. Or... \(\displaystyle x^6-y^6 = (x^2)^3-(y^2)^3\)
you have the right idea. \(\displaystyle x^6-y^6 = (x^3)^2-(y^3)^2\) there's your difference of two squares. Then you have difference and sum of two cubes. Or... \(\displaystyle x^6-y^6 = (x^2)^3-(y^2)^3\)
M masters Full Member Joined Mar 30, 2007 Messages 378 Mar 5, 2010 #3 gemini77 said: factoring x^6 - y^6 difference of squares is (x-y)(x+y) X^3-y^3 is (x-y)(x^2 + xy + y^2) to ^4 is (x-y)(x+y) (x^2 + xy + y^2) what is to the ^5 or ^6 look like?????? Click to expand... Hi gemini, First, factor as the difference of two squares. Then, factor the sum and difference of two cubes. \(\displaystyle x^6-y^6=(x^3)^2-(y^3)^2=(x^3+y^3)(x^3-y^3)=(x+y)(x^2-xy+y^2)(x-y)(x^2+xy+y^2)\)
gemini77 said: factoring x^6 - y^6 difference of squares is (x-y)(x+y) X^3-y^3 is (x-y)(x^2 + xy + y^2) to ^4 is (x-y)(x+y) (x^2 + xy + y^2) what is to the ^5 or ^6 look like?????? Click to expand... Hi gemini, First, factor as the difference of two squares. Then, factor the sum and difference of two cubes. \(\displaystyle x^6-y^6=(x^3)^2-(y^3)^2=(x^3+y^3)(x^3-y^3)=(x+y)(x^2-xy+y^2)(x-y)(x^2+xy+y^2)\)
