There's a pre-calculus test tomorrow and I'm lost

All Nines

Could anyone help me on one more before I go to bed and hope for the best?

csc^6(x) - cot^6(x) + 1 = 3(cot^2(x)csc^2(x))

If I've taken the correct steps (using difference of cubes), I have (deep breath):

(csc^2(x) - cot^2(x)) * (csc^4(x) + csc^2(x)cot^2(x) + cot^4(x)) + 1 = 3(cot^2(x)csc^2(x))

But at that point I'm lost.

Ponicalica

Multiply both sides by sin^6(x).

All Nines

As I said above, I can only manipulate one side of the equation.

But just because I'd like to know, how does that work?

Stormtroper

Differential equations: The way things change change the way it changes.

Chagen: I've learned that it's not so much the formulas themselves, though they definitely have many applications, but it's the brain-practice of solving problems by looking at the pieces and then applying the right tools (such as math formulas) to put them together.

So very much this.

FrodoGoofballCoTV

^agreed.

^x4 http://answers.yahoo.com/question/index?qid=20101103001418AAJqvPX

glennmagusharvey

Everest: If you can't manipulate both sides, then just multiply and divide by the same thing.