Identity problem #2: sin^3 - 8/sin - 2 = sin^2 + 2sin + 4

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Identity problem #2: sin^3 - 8/sin - 2 = sin^2 + 2sin + 4

Mathematics
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sin^4-8-2sin=sin^2+2sin+4.....
subtract
wait....nevermind these are proofs not equations...give me a minute.

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Okay, take your time. XD
this is a tough one......
they are fun though..give a minute....
sin^3 - 8 -------- = sin^2 + 2sin + 4 ?? is this right? sin - 2
the top is a difference of cubes.....
(sin-2) (sin^2 +2sin +4) ------------------- (sin-2)
really? i dont see that?
sin^3 - 2^3.... plain as day :)
yeah, but what about 1/sin?
there is no 1/sin.... which problem you doin? :)
same one, 8/sin
nah.... that aint it.....
(sin^3 - 8)/(sin - 2) = sin^2 + 2sin + 4
there are no parentheses in the problem, so im assuming that it means -8/sin
then it aint gonna work out :) Have a crack at it, but its futile.....
It's (sin^3 - 8)/(sin-2)
oh.....ok. You were right. My bad. sorry.
'sok :) Blame it on the rain :)
cool.
How does a difference of cubes work? Can you show me it in detail?
The best way to understand it is to actually multiply a cubed expression like (a-b)^3
Which of course is just (a-b)^2 times (a-b)
Ack!!...splained it wrong lol (a-b)(a^2 + ab + b^2)
suppose you have an expression like this (f = first and l = last): (f^3 - l^3) can be factored: (f-l)(f^2 +fl +l^2)
So what would be a and b? I'm sorry, I'm no good at this stuff. Thanks for helping though!
Lets use some numbers: x^2 - 64 is a good example: 4*4*4 = 64 = 4^3 x^3 - 4^3 can be factored like this: (x-4)(x^2 +4x + 4^2)
that should have been x^3 - 64....
many centuries ago, mathmatickers noticed patterns in the way certain forms worked; and they surmised these tricks to help them solve the stuff quicker....
Wow, I get it now! Thanks soooo much!!
good job... I gotta be going home now.... Ciao :)
Night!

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