# 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?

• Let $a = 1$ and $b=-1$. – Jacky Chong Oct 5 '16 at 21:50
• I think you have the impose the condition $a, b\geq 0$ or $a, b \leq 0$. – Jacky Chong Oct 5 '16 at 21:51
• @SimpleArt I did not understand you. If $a=b=1$ or $a=\sqrt {b}$ then, what can it be? – James Ensor Oct 5 '16 at 21:55
• @Kahler No, I just read the problem wrong. Anyways, all the other answers are fine. – Simply Beautiful Art Oct 5 '16 at 21:55
• @JackyChong You are right. I should disprove. – James Ensor Oct 5 '16 at 21:56

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.
• I am thinking you should restrict $a\times b\ge0$, this allows $a,b$ to be both negative. – Simply Beautiful Art Oct 5 '16 at 21:56
• @Kahler No, it implies $a=0$ or $b=0$: $$\begin{cases}a=0\\b=0\\a=-b\end{cases}$$ – Simply Beautiful Art Oct 5 '16 at 22:00
• @Kahler You can't prove that, because it isn't true. The best you can say is that $a=0$ or $b=0$ or $a=-b$. – wgrenard Oct 5 '16 at 22:02
• Come up with a counterexample. Find two real numbers $a$ and $b$ where $(a+b)^3=a^3+b^3$ but $a$ and $b$ are not both zero. This will disprove the statement. Hint: let $a=0$ and $b$ equal anything else besides zero. – wgrenard Oct 5 '16 at 22:04