how to factor this ? 1+y^12 =

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

how to factor this ? 1+y^12 =

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

that is a sum of cubes \[a^3+b^3=(a+b)(a^2-ab+b^2)\] Where in your expression \(a=1\) and \(b=y^4\)
thanks :)
@peachpi how to determine faster if square or cube ?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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 square first 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
thanks for your time :)

Not the answer you are looking for?

Search for more explanations.

Ask your own question