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.
 A: Hint:$$\left( \sin  a+\cos  a \right) \left( \sin ^{ 2 } a-\sin  a\cos  a+\cos ^{ 2 } a \right) =\\ =\left( \sin  a+\cos  a \right) \left( { \left( \sin  a+\cos  a \right)  }^{ 2 }-\frac { 3 }{ 2 } \left( { \left( \sin { a } +\cos { a }  \right)  }^{ 2 }-1 \right)  \right)  $$
A: HINT: You know what $\sin^2\alpha+\cos^2\alpha$ is, so the problem boils down to sorting out $\sin\alpha\cos\alpha$. But
$$(\sin\alpha+\cos\alpha)^2=\sin^2\alpha+2\sin\alpha\cos\alpha+\cos^2\alpha\;,$$
and you can solve this for $\sin\alpha\cos\alpha$.
A: Hint: if $\sin\alpha+\cos\alpha=1.2$ then 
$$
(\sin\alpha+\cos\alpha)^2=\sin^2\alpha+2\sin\alpha\cos\alpha+\cos^2\alpha=1+2\sin\alpha\cos\alpha=1.44.
$$
A: $$\sin^3 \alpha+\cos^3 \alpha=(\sin \alpha+\cos \alpha)(\sin^2 \alpha-\sin \alpha\cos \alpha+\cos^2 \alpha)$$
$$1.2^3=\sin^3\alpha+\cos^3 \alpha+3\sin \alpha\cos \alpha(\sin \alpha+\cos \alpha)$$ It follows $$\sin \alpha\cos \alpha=0.22$$ and because $$\sin \alpha+\cos \alpha=1.2$$ one has sum and product of two unknowns so its are solution of the quadratic equation $$X^2-1.2X+0.22=0$$ with positive discriminant $0.14$.
A: ${x^3+y^3 \over x+y} = x^2+y^2-xy$. 
$(x+y)^2 = x^2+y^2+ 2 xy$, 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 $x^3+y^3 = (1.2)(1-{1 \over 2}((1.2^2-1)) = {117 \over 125}$.
