Answer:
1
Step-by-step explanation:
Use the form for the factoring of the sum of cubes.
To make this easier to write, we'll use the substitutions ...
a = sin(θ), b = cos(θ)
It looks like you may want to simplify ...
(a³ +b³)/(a +b) +ab
= ((a +b)(a² -ab +b²))/(a +b) +ab
The factors of (a+b) cancel. In this case, we know that a² +b² = 1, so this becomes ...
= (1 -ab) +ab
= 1
[tex]\boxed{\dfrac{\sin^3{\theta}+\cos^3{\theta}}{\sin{\theta}+\cos{\theta}}+\sin{\theta}\cos{\theta}=1}[/tex]
