$${ \sin }^{ 3 } 2 \pi /9- { \sin }^{ 3 } \pi /9$$
$0$
$$\frac{0^{3}}{9}-\frac{\left(\sin(\pi )\right)^{3}}{9}$$
$$\frac{0}{9}-\frac{\left(\sin(\pi )\right)^{3}}{9}$$
$$0-\frac{\left(\sin(\pi )\right)^{3}}{9}$$
$$0-\frac{0^{3}}{9}$$
$$0-\frac{0}{9}$$
$$0+0$$
$$0$$
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