MHB Which method is easier and why?

AI Thread Summary
The discussion focuses on factoring the expression x^6 - y^6 using two methods: difference of squares and difference of cubes. It is noted that applying the difference of squares leads to simpler subsequent steps, resulting in the factorization x^6 - y^6 = (x^3 + y^3)(x^3 - y^3). This method further breaks down into (x + y)(x^2 - xy + y^2)(x - y)(x^3 + xy + y^2). In contrast, using the difference of cubes yields a more complex factorization, although it is still manageable. Overall, the difference of squares is recommended for its simplicity in the next steps.
mathdad
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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?
 
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Excellent!
 
RTCNTC said:
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
 
kaliprasad said:
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.
 
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