MHB Which method is easier and why?

[mathdad]

I found the following problem in Section 1.3 of my Precalculus textbook by David Cohen.

Factor x^6 - y^6 as a difference of squares and then as a difference of cubes.

1. Which method is easier and why?

2. Can someone get me started?

 

[joypav]

View attachment 6492

 

[mathdad]

Excellent!

 

[kaliprasad]

I found the following problem in Section 1.3 of my Precalculus textbook by David Cohen.

Factor x^6 - y^6 as a difference of squares and then as a difference of cubes.

1. Which method is easier and why?

2. Can someone get me started?

you can apply either way. but next steps become simpler if you apply difference of square

$x^6-y^6= (x^3+y^3)(x^3-y^3)$ 1st term is sum of cubes and second is difference of cubes and you get

$x^6-y^6= (x^3+y^3)(x^3-y^3)= (x+y)(x^2-xy+y^2)(x-y)(x^3+xy+y^2)$

had you chosen difference of cubes you would have got

$x^6-y^6 = (x^2-y^2)(x^4+x^2y^2+y^4)$

factoring $x^4+x^2y^2+y^4 = (x^4+2x^2y^2+y^4)- x^2y^2 = (x^2+y^2)^2 - (xy)^2 = (x^2+y^2+xy)(x^2+y^2-xy)$

above may not be straight forward but doable

 

[mathdad]

you can apply either way. but next steps become simpler if you apply difference of square

$x^6-y^6= (x^3+y^3)(x^3-y^3)$ 1st term is sum of cubes and second is difference of cubes and you get

$x^6-y^6= (x^3+y^3)(x^3-y^3)= (x+y)(x^2-xy+y^2)(x-y)(x^3+xy+y^2)$

had you chosen difference of cubes you would have got

$x^6-y^6 = (x^2-y^2)(x^4+x^2y^2+y^4)$

factoring $x^4+x^2y^2+y^4 = (x^4+2x^2y^2+y^4)- x^2y^2 = (x^2+y^2)^2 - (xy)^2 = (x^2+y^2+xy)(x^2+y^2-xy)$

above may not be straight forward but doable

Nice work in every sense.