= (csc^2(x) - cot(x) - 2cos^2(x))/1 - cot^2(x)

Post your question with the appropriate groupings making use of brackets please and avoid omitting signs. The way you gave us is this way:


\frac{(sin^3(x) cos^3(x))}{sin(x)} cos(x)
= \frac{(csc^2(x) - cot(x) - 2cos^2(x))}{1} - cot^2(x)

Which is not true.

Sorry I wrote the question wrong this is the correct one, verify this identity: (sin^3(x) cos^3(x)) /(sin(x) + cos(x)) = (csc^2(x) - cot(x) - 2cos^2(x)) / (1 - cot^2(x))

Sorry I wrote this question wrong again this is really the correct one, verify this identity: (sin^3(x) - cos^3(x)) / (sin(x) + cos(x)) = (csc^2(x) - cot(x) - 2cos^2(x)) / (1 - cot^2(x))

Okay.

\frac{sin^3(x) - cos^3(x)}{sin(x) + cos(x)} = \frac{csc^2(x) - cot(x) - 2cos^2(x)}{1 - cot^2(x)}

I think it would be easier to start with the left part:

\begin{array}{rcl}

\frac{sin^3(x) - cos^3(x)}{sin(x) + cos(x)} &=& \frac{(sin(x) - cos(x))(sin^2(x) + sin(x)cos(x) + cos^2(x))}{sin(x) + cos(x)}\\

&=& \frac{(sin(x) - cos(x))^2(1 + sin(x)cos(x))}{(sin(x) + cos(x))(sin(x) - cos(x))}\\

&=& \frac{(sin^2(x) + cos^2(x) -2sin(x)cos(x))(1 + sin(x)cos(x))}{sin^2(x) - cos^2(x)}\\

&=& \frac{1 -2sin(x)cos(x))(1 + sin(x)cos(x))}{sin^2(x) - cos^2(x)}\\

&=& \frac{1 - cos(x)sin(x) - 2cos^2(x)sin^2(x)}{sin^2(x) - cos^2(x)}\\

&=& \frac{csc^2(x) - cot(x) - 2cos^2(x)}{1 - cot^2(x)}\\
\end{array}

1. Factorise in the form a^3 - b^3 = (a - b)(a^2 + ab + b^2).
2. Multiply by (sin(x) - cos(x)) in both numerator and denominator.
3. Expand and simplify numerator and denominator.
4. Simplify with identity cos^2(x) + sin^2(x) = 1
5. Expand numerator.
6. Divide numerator and denominator by sin^2(x)