Find all Integral solutions to $x+y+z=3$, $x^3+y^3+z^3=3$. Suppose that $x^3+y^3+z^3=3$ and $x+y+z=3$. 

What are all integral solutions of this equation? 

I can only find $x=y=z=1$.
Using identity: $(x+y+z)^3 = x^3+y^3+z^3 + 3(x+y)(y+z)(z+x)$, we have:

$3^3 = 3 + 3(x+y)(y+z)(z+x) \Rightarrow (x+y)(y+z)(z+x) = 8$. From this you should be able to deduce the answer.
Using the Power Mean Inequality we have that:

$$ 1=\frac{1}{3}(x^3 + y^3 + z^3)  \leq \left[\frac{1}{3}(x+y+z)\right]^3 =1$$

Equality only holds when $x=y=z=1$.  



Let's replace $z \mapsto -z$ and try to solve the system of equations for $x,y,z \geq 0$:

$$ x^3 + y^3 = 3 + z^3 \text{ and } x + y = z + 3 $$

We can still use the power mean inequality (in only two of the varaibles) and use the linear relation

$$ \frac{1}{3}z^3 &lt; \frac{1}{3}(3 + z^3 ) = \frac{1}{3}(x^3 + y^3 )  \leq \left[\frac{1}{3}(x+y)\right]^3 = \left[\frac{1}{3}(z+3)\right]^3
&lt; \frac{1}{24}(z+3)^3 $$

Maybe if I take cube roots the left and right sides are easier to compare.  I also rounded $3^3 \approx 27$

$$ z &lt;  \frac{1}{2}(z + 3)$$

This means $0 \leq z &lt; 3$.  Can it be that $0^3 + 3 = 3, 1^3 + 3 = 4, 2^3 + 3 = 11$ are the sums of two cubes?



If two of the 3 numbers are negative,  the starting point should be:

$$ x^3 + y^3 = -3 + z^3 \text{ and } x + y = z - 3 $$

Then using the power mean inequality we can attempt to deduce a range on $z$

$$  \frac{1}{3}(-3 + z^3 ) &lt; \frac{1}{24}(z-3)^3 $$

This one is harder to estimate.  After fiddling around, we see that $z \leq 0$:

$$   z &lt;  \frac{1}{2}\sqrt[3]{(z-3)^3 + 24} 
&lt; \frac{1}{2}z $$

So there are definitely no solutiuons of this kind.