Let $x, y \in [0, 2 π]$ and satisfying the equation $\sin^{3} x + \cos^{3} y + 6 \sin x . \cos y = 8$
If $l = x + y, m = x - y$ then

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least value of $l$ is $\frac{π}{2}$

least value of $m$ is $- \frac{3 π}{2}$

largest value of $l$ is $\frac{5 π}{2}$

largest value of $m$ is $\frac{π}{2}$

answer is A, B, C, D.

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Detailed Solution

$a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a)$
$If a + b + c = 0, then a^{3} + b^{3} + c^{3} = 3 a b c$
$Now, \sin^{3} x + \cos^{3} y + (- 2)^{3} = 3 \sin x \cdot \cos y \cdot (- 2)$
$\begin{array}{l} \Rightarrow \sin x + \cos y - 2 = 0 \\ \Rightarrow \sin x + \cos y = 2 \\ \Rightarrow \sin x = 1, \cos y = 1 \\ \Rightarrow x = \frac{π}{2}, y = 0, 2 π \\ ∴ l = x + y = \frac{π}{2}, \frac{5 π}{2} \\ ∴ m = x - y = \frac{π}{2}, - \frac{3 π}{2} \end{array}$