Hello Hello,
We are given to prove:
$$\frac{\sin^4(x)-\cos^4(x)}{\sin^3(x)+\cos^3(x)}=\frac{\sin(x)-\cos(x)}{1-\sin(x)\cos(x)}$$
Traditionally, we begin with the left side and try to apply well-known algebraic formulas and trigonometric identities to obtain the right side. So let's look at the left side:
$$\frac{\sin^4(x)-\cos^4(x)}{\sin^3(x)+\cos^3(x)}$$
Now, it we factor the numerator as the difference of squares, and the denominator as the sum of cubes, we may write:
i) Difference of squares:
$$a^2-b^2=(a+b)(a-b)$$
ii) Sum of cubes:
$$a^3+b^3=(a+b)\left(a^2-ab+b^2 \right)$$
$$\frac{\sin^4(x)-\cos^4(x)}{\sin^3(x)+\cos^3(x)}=\frac{\left(\sin^2(x)+\cos^2(x) \right)\left(\sin^2(x)-\cos^2(x) \right)}{\left(\sin(x)+\cos(x) \right)\left(\sin^2(x)-\sin(x)\cos(x)+\cos^2(x) \right)}$$
Applying the Pythagorean identity $$\sin^2(\theta)+\cos^2(\theta)=1$$ we have:
$$\frac{\sin^4(x)-\cos^4(x)}{\sin^3(x)+\cos^3(x)}=\frac{\sin^2(x)-\cos^2(x)}{\left(\sin(x)+\cos(x) \right)\left(1-\sin(x)\cos(x) \right)}$$
Factoring the numerator as the difference of squares, we find:
$$\frac{\sin^4(x)-\cos^4(x)}{\sin^3(x)+\cos^3(x)}=\frac{\left(\sin(x)+\cos(x) \right)\left(\sin(x)-\cos(x) \right)}{\left(\sin(x)+\cos(x) \right)\left(1-\sin(x)\cos(x) \right)}$$
Dividing out common factors, we see:
$$\frac{\sin^4(x)-\cos^4(x)}{\sin^3(x)+\cos^3(x)}= \frac{\cancel{\left(\sin(x)+\cos(x) \right)} \left(\sin(x)-\cos(x) \right)}{\cancel{\left(\sin(x)+ \cos(x) \right)}\left(1-\sin(x)\cos(x) \right)}$$
$$\frac{\sin^4(x)-\cos^4(x)}{\sin^3(x)+\cos^3(x)}=\frac{\sin(x)-\cos(x)}{1-\sin(x)\cos(x)}$$
Shown as desired.