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?

1 Answers1

4

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.

wgrenard
  • 3,678
  • 1
    I am thinking you should restrict $a\times b\ge0$, this allows $a,b$ to be both negative. – Simply Beautiful Art Oct 05 '16 at 21:56
  • @SimpleArt yep, that works too, and it's more general. Nice observation. – wgrenard Oct 05 '16 at 21:58
  • 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?

     –  Oct 05 '16 at 21:58
  • 1
    @Kahler No, it implies $a=0$ or $b=0$: $$\begin{cases}a=0\b=0\a=-b\end{cases}$$ – Simply Beautiful Art Oct 05 '16 at 22:00
  • 1
    @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 05 '16 at 22:02
  • so, how can I disprove my last question (for ''and'')? –  Oct 05 '16 at 22:02
  • 1
    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 05 '16 at 22:04
  • @wgrenard Let $a=0$ and $b=2$. It is a counterexample, ,isn't it? –  Oct 05 '16 at 22:06
  • 1
    @Kahler You are correct! – wgrenard Oct 05 '16 at 22:07
  • @wgrenard in addition, let $f$ and $g$ to be two functions from $\mathbb{R}$ to $\mathbb{R}$. If for all $x$ the equation $f\left( x\right) g\left( x\right)=0$ holds then either $f\left( x\right)=0$ for all $x$ or $g\left( x\right)=0$ for all $x$. it is a counterexample: $f\left( x\right)=0$ and $g\left( x\right)=0$ for all $x$, isn't it? –  Oct 05 '16 at 22:27
  • 1
    @Kahler No, that is not a counterexample because in that example the statement "$f(x)=0$ for all $x$ or $g(x)=0$ for all $x$" certainly holds. You need to find an example where $f(x)$ is not identically zero and $g(x)$ is not identically zero. If you still have questions after thinking about that for awhile, I recommend you post a new question along with your thoughts on the problem, as the comment section is not meant for discussion :) – wgrenard Oct 05 '16 at 22:32
  • okey, thanks... –  Oct 05 '16 at 22:43