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that is a sum of cubes
Where in your expression \(a=1\) and \(b=y^4\)
@peachpi how to determine faster if square or cube ?
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For a difference I'm not sure it matters because you will end up with the same solution either way. For a sum, you can only factor a sum of cubes with involving complex numbers.
the exponent of the variable ? like 12 ? how do we know faster to determine square or cube ? how about like y^18 y^24 ?
For exponents divisible by both 2 and 3 the exponents themselves don't matter, it's the sign between the terms that does.
1+y^12, 1+y^18 and 1+y^24 you have to factor using cube because there's no rule for a sum of squares.
1-y^12, 1 - y^18, and 1 - y^24 I believe you should be able to apply either the difference of cubes or difference of squares rule and get the same result. I personally prefer to do the square factor first because then you don't have to deal with the trinomial
1 - y^12
(1 + y^6)(1-y^6)
For (1 + y^6) you have to use the sum of cube rule. For (1-y^6) you can again choose either rule. I'll do square to be consistent
(1 + y^2)(1 - y^2 + y^4)(1+y^3)(1-y^3)
The first two are irreducible, the second two are no longer squares, so they must be factored using as cubes
(1 + y^2)(1 - y^2 + y^4)(1+y)(1 - y + y^2)(1 - y)(1 + y + y^2)
Try doing the cube first and you should get the same result