Question

Can anybody help me with this?

Answer 1

\[x^{12} - 64 = (x ^{6})^{2} - 8^{2}\]
Which is a difference of two squares.
(x^6 + 8)(x^6 - 8)
Now we have sum of two cubes!
Remember that
\[A^3 + B^3 = (A + B)(A^2 -AB +B^2)\] and \[A^3 - B^3 = (A - B)(A^2 +AB +B^2)\]
So let A = x^2 and B = 2.
\[x^6 + 8 = (x^2)^3 + 2^3 = (x^2 + 2)(x^2 -2x^2 +4)\]
\[x^6 - 8 = (x^2)^3 + 2^3 = (x^2 - 2)(x^2 +2x^2 +4)\]

Answer 2

Thank you so much @mww ! Do I have to continue something or how? I am sorry but I am really having problems with the factoring :(

Answer 3

Multiply the last two together to get D.
All you do is figure out which factoring method is best
We simplified it using diff. of 2 squares
Then saw that sum and diff of 2 cubes worked best for the next step.
I showed you the sum of 2 cubes and diff of 2 cubes separately. All you do is multiply those together.

Answer 4

Thank you!!!

Answer 5

@mww do you mind helping with one more of these? I can't seem to factor it:(

Answer 6

sure