Modify it your way!

Using the identity $a^{3} + b^{3} = (a + b)(a^{2} - ab + b^{2})$ with in this case $a = \sin^{2}(x)$ and $b = \cos^{2}(x)$ , along with the identity $\sin^{2}(x) + \cos^{2}(x) = 1$ , we have that

$\sin^{6}(x) + \cos^{6}(x) = (\sin^{2}(x) + \cos^{2}(x))(\sin^{4}(x) - \sin^{2}(x)\cos^{2}(x) + \cos^{4}(x)) =$

$(\sin^{2}(x) + \cos^{2}(x))^{2} - 3\sin^{2}(x)\cos^{2}(x) = 1 - \dfrac{3}{4} \times (2\sin(x)\cos(x))^{2} = 1 - \dfrac{3}{4}\sin^{2}(2x)$ ,

since $\sin(2x) = 2\sin(x)\cos(x)$ . Now $-1 \le \sin(2x) \le 1 \Longrightarrow 0 \le \sin^{2}(2x) \le 1$ , so

$1 - \dfrac{3}{4} \le 1 - \dfrac{3}{4}\sin^{2}(2x) \le 1$ , and thus $A + B = \dfrac{1}{4} + 1 = \dfrac{5}{4} = \boxed{1.25}$ .

Differentiate the equation into 6sin(x)cos(x) * [ $sin^{4}$ (x)- $cos^{4}(x)$ ]

Then set it equal to zero for the x coordinates of the extrema.

Maximum values is 1.

Minimum value is 0.25.

Add them together for the answer of $\boxed{1.25}$ .