81s^6-3f^6 Factor

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81s^6-3f^6 Factor

Mathematics
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factor out the 3 and look for diff of squares
or even diff of cubes
27 is a cube so...

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Other answers:

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

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