Question

the complete factored form of y^8-256 is??

Answer 1

This expression is the difference of two cubes. Use the formula.

Answer 2

y^8 - 256 can not be factorized further since there is no factor common of given expression

Answer 3

256 is a perfect square, not a cube. I have to agree with kushashwa.

Answer 4

thanks if u agree than give medal

Answer 5

(y-2)(y+2)(y^2+4)(y^4+16) are the factors

Answer 6

i am definitely sure my answer is correct

Answer 7

yes yes i understood that u r corret aryan very gud 1 medal from my side

Answer 8

thanks man

Answer 9

what is the fornula?

Answer 10

a differenece of squares is (y^2-b^2) = (y+b) (y-b) but in this case it's starting off w/ a squared square (y^4-b^4) so it's a difference of fourths (y^2+b^2) (y^2-b^2) but this second part breaks down to (y+b) (y-b) so it's (y^2+b^2) (y+b) (y-b)

(y^2 + 16) (y^2 - 16)

(y^2 + 16) (y + 4)(y- 4)

Answer 11

\[y^8-256=(y^4-16)(y^4+16)=(y^2-4)(y^2+4)(y^4+16)=\]
\[=(y-2)(y+2)(y^2+4)(y^4+16)\]

Answer 12

\[(y-2)(y+2)(y^2+4)(y^4+16)=\]
\[=(y-2)(y+2)(y^2+4)(y^2-2\sqrt{2}y+4)(y^2+2\sqrt{2}y+4)\]

Answer 13

oh my gosh. thank you so much.