<s>[INST] If $\sin \alpha +\cos \alpha =1.2$, then what is $\sin^3\alpha + \cos^3\alpha$? If $\sin \alpha +\cos \alpha =1.2$, then what is $\sin^3\alpha + \cos^3\alpha$?

All I know is that $\sin^{3}a+\cos^{3}a$ is equal to 
$$(\sin a + \cos a)(\sin^{2} a - \sin a \cos a + \cos^{2} a)= \dfrac{6}{5}(\sin^{2} a - \sin a \cos a + \cos^{2} a)$$  But now, I'm stuck.  Solutions are greatly appreciated.
 [/INST] ${x^3+y^3 \over x+y} = x^2+y^2-xy$. 
$(x+y)^2 = x^2+y^2+ 2 xy$, and so
${x^3+y^3 \over x+y} = x^2+y^2  -{1 \over 2} ((x+y)^2-(x^2+y^2))$, from which we get
$x^3+y^3 = (x+y) (x^2+y^2  -{1 \over 2} ((x+y)^2-(x^2+y^2)))$.
Since $x+y = 1.2, x^2+y^2 = 1$ we get $x^3+y^3 = (1.2)(1-{1 \over 2}((1.2^2-1)) = {117 \over 125}$.
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