I needd helppp. I can't verify this identity! Pleasee help and show full work. Thanks :)

Ok, you need to find some way to break down the 'cubes' in the cos and sin functions.

Separate the cos^3 and sin^3 into cos(x)cos^2(x) and sin(x)sin^2(x) respectively:

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

Use the identity cos^2(x) + sin^2(x) = 1:

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

Expand;

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

Rearrange;

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

Factor out cos(x)sin(x) from the two last terms in the numerator;

= \frac{(sin(x)+ cos(x)) - (cos(x)sin(x))(sin(x)+cos(x))}{sin(x) + cos(x)}

Factor out sin(x) + cos(x) from the numerator in all the terms together;

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

It's done now. :) Only one little step more.

If I may show another way.

Note the sum of two cubes in the numerator. Factoring it, gives:

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

The sin(x)+cos(x) cancels and we are left with:

sin^{2}(x)-sin(x)cos(x)+cos^{2}(x)

Note the identity sin^{2}(x)+cos^{2}(x)=1

This leaves us with

1-sin(x)cos(x)