Question

81s^6-3f^6
Factor

Answer 1

factor out the 3 and look for diff of squares

Answer 2

or even diff of cubes

Answer 3

27 is a cube so...

Answer 4

3(27s^6 - f^6)

3 (3s^2 -f^2) (9s^4 +3(sf)^2 +f^4) if I remember correctly; and then theres that diff of squares that it generated....

Answer 5

you have any with sqrt(3) in it?

Answer 6

3 (sqrt(3)s +f)(sqrt(3)s -f) (9s^4 +3(sf)^2 +f^4) but let me check thru that again

Answer 7

not they all have ^6

Answer 8

all the answers?

Answer 9

3(3s^2-f^2)(9s^4+3f^2s^2+f^4)

Answer 10

3(27s^6 -f^6) is what it initially factor to

Answer 11

but thats a difference of cubes

Answer 12

ok so what shall I put in for the answer then please

Answer 13

\[3(3s^2-f^2)(9s^4+3f^2s^2+f^4)\] is the answer

Answer 14

if we factor this completely i get:

3 .... yeah, like that again

Answer 15

but, that (3s^2 -f^2) factors again

Answer 16

so do u amistre64 agree with smurfy answer

Answer 17

\[3(\sqrt{3}+f)(\sqrt(3)-f)(9s^4 +3s^2 f^2 +f^2)\]

Answer 18

3(3s2−f2)(9s4+3f2s2+f4)

Answer 19

that (3) is a typo; just sqrt(3) like the other one lol

Answer 20

This is the answer: 3(3√+f)((√3)−f)(9s4+3s2f2+f2)

Answer 21

i just dont know what your material is looking for for an acceptable answer with that

Answer 22

\(\sqrt{3}\) is a valid number as far as i can tell

Answer 23

it just say factor sir

Answer 24

I think i'll learn partial differential equations before she learns how to factor

Answer 25

my gut says to go all the way ;)

\(3(\sqrt{3}+s)(\sqrt{3}-s)(9s^4+3s^2 f^2+f^4)\)

Answer 26

ok i will try it

Answer 27

ack..cant type lol

Answer 28

lordamercy. isn't this just the difference of two cubes?

Answer 29

make it:

\(3(\sqrt{3}s-f)(\sqrt{3}s+f)(9s^4 +3s^2 f^2 +f^4)\)

Answer 30

thats my final offer lol

Answer 31

its a diff of cubes with a diff of squares snucked in

Answer 32

im telling you mine is correct you can check it if you dont believe me

Answer 33

3(3√+f)((√3)−f)(9s4+3s2f2+f2)

Answer 34

checking a partial simplification doesnt prove nuthin lol

Answer 35

amistre64 is smarter!

Answer 36

at times.... other times im just an idiot in disguise ;)

Answer 37

haha

Answer 38

ok smurfy you were right sorry

Answer 39

thank you :)

Answer 40

yay!! smurfy

Answer 41

even tho i typed it in like that first ...... lol

Answer 42

\[3(27s^6-f^6)=3((s^2)^3-(f^2)^3\]\]
\[a^3-b^3=(a-b)(a^2+ab+b^2)\]
so you get \[3((3s^2)-f^2)((3s^2)^2+3s^2f^2+(f^2)^2)\]
\[3(3s^2-f^2)(9s^4+3s^2f^2+f^4)\]

Answer 43

hello amistre!

Answer 44

yes; and the question I had was do we take the diff of squares further?

Answer 45

howdy! :)

Answer 46

ahh i see . first term is difference of two squares, if you are factoring over reals. yes.

Answer 47

thats what I thought .....