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factor out the 3 and look for diff of squares

or even diff of cubes

27 is a cube so...

you have any with sqrt(3) in it?

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

not they all have ^6

all the answers?

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

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

but thats a difference of cubes

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

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

if we factor this completely i get:
3 .... yeah, like that again

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

so do u amistre64 agree with smurfy answer

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

3(3s2−f2)(9s4+3f2s2+f4)

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

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

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

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

it just say factor sir

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

my gut says to go all the way ;)
\(3(\sqrt{3}+s)(\sqrt{3}-s)(9s^4+3s^2 f^2+f^4)\)

ok i will try it

ack..cant type lol

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

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

thats my final offer lol

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

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

3(3√+f)((√3)−f)(9s4+3s2f2+f2)

checking a partial simplification doesnt prove nuthin lol

amistre64 is smarter!

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

haha

ok smurfy you were right sorry

thank you :)

yay!! smurfy

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

hello amistre!

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

howdy! :)

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

thats what I thought .....