Q:

Let $a$ and $b$ be real numbers. Then $\left( a+b\right) ^{3}=a^{3}+b^{3}$ implies $a=0$ or $b=0$.

My proof. We need to show that $a=0$ or $b=0$ for the equation. 
    We have,
$\left( a+b\right) ^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3}$ (by the binomial theorem)
$=a^{3}+b^{3}$ (by the assumption).
Now, adding $(-(a^{3}+b^{3})$ both sides yields $3a^{2}b+3ab^{2}=0$, i.e., $3ab\left( a+b\right) =0$ (by the distributive law). Since $3ab\left( a+b\right) =0$, $3ab=0$ or $(a+b)=0$, i.e., $a=0$ or $b=0$.
Can you check my proof?
in addition, Let $a$ and $b$ be real numbers. Then $\left( a+b\right) ^{3}=a^{3}+b^{3}$ implies $a=0$ and $b=0$.
So, how can I show this?

A:

What you are trying to prove is not true. For instance, let $a=1$ and $b=-1$, then $(a+b)^3=a^3 + b^3=0$.
The final conclusion of your attempted proof is incomplete, because $3ab(a+b) = 0$ implies that $a=0$ or $b=0$ or $a=-b$.
The case where $a=-b$ can be eliminated if you restrict $a$ and $b$ to be nonnegative. Thus, if $a,b \geq 0$ your proof is correct.

