Question

Okay, is this factoring right?

Answer 1

\[64-z ^{6}\]
\[(8^{2})-(z ^{4})^{6}\]

Answer 2

\[\huge{a^2-b^2=(a+b)(a-b)}\]
\[\huge{\text{Use this identity}}\]

Answer 3

\[\huge{8^2-(z^3)^2}\]

Answer 4

4 times 6 does not equal 6

Answer 5

Oh, still getting the hang of this..

Answer 6

mathslover is correct

Answer 7

\[\Huge{(8-z^3)(8+z^3)}\]
\[\Huge{(2^3-z^3)(2^3+z^3)}\]

Answer 8

we made 8 as 2^3

Answer 9

now we will apply \(\huge{a^3-b^3=(a-b)(a^2+ab+b^2)}\)

Answer 10

and \(\huge{a^3+b^3=(a-b)(a^2+ab+b^2)}\)

Answer 11

got it?

Answer 12

I did, I think the asker left