Question on convergence of difference between two real sequences. 
Let $\{x_n\}_{n \geq 1}$ and $\{y_n\}_{n \geq 1}$ be two sequences of real numbers such that $x_n^3 - y_n^3 \to 0.$ Can we always conclude that $x_n - y_n \to 0\ $?

I think it is true. But I can't able to prove it. I tried by factorizing $$x_n^3 - y_n^3 = (x_n - y_n) (x_n^2 + x_n y_n + y_n^2).$$
Now how do I proceed? Any help will be appreciated.
Thanks!
 A: Yes.

For $x, y \in \Bbb R$, we have $$x^2 + xy + y^2 \ge \frac{3}{4}y^2.$$
(Complete the square.)
Let $c := \frac{3}{4}$.
Thus, we get $$|x^3 - y^3| \ge c|x - y|y^2 \ge 0.$$
Now, since $x_n^3 - y_n^3 \to 0$, Sandwich Theorem tells us that $$|x_n - y_n|y_n^2 \to 0$$ and the same with $x_n$ instead of $y_n$.
To conclude, we get
$$|x_n - y_n|(x_n^2 + y_n^2) \to 0. \qquad (*)$$
Now, note that $$\sqrt{2}\sqrt{x_n^2 + y_n^2} \ge |x_n| + |y_n| \ge |x_n - y_n|$$
and thus, plugging it in $(*)$ gives
$$0 \le \frac{1}{2}|x_n - y_n|^3 \le |x_n - y_n|(x_n^2 + y_n^2) \to 0.$$
Sandwich theorem now gives gives $$|x_n - y_n|^3 \to 0.$$
Thus, $$|x_n - y_n| \to 0,$$ as desired.
A: Consider $\displaystyle x_n^3 - y_n^3 = \int\limits_{y_n}^{x_n} dt\ \dfrac{t^2}{3},$ this converges to zero in absolute value, so
$$
|x_n^3 - y_n^3| \geq \frac{1}{3}\min(x_n^2, y_n^2) |x_n - y_n|.
$$
If either $x_{n_k}^3$ or $y_{n_k}^3$ converges to any number $a^3$ (for some subsequence $(n_k)$), then $x_{n_k}$ and $y_{n_k}$ will both converge to $a$ by the continuity of the cubic root. Further, we know that a sequence converges if and only every subsequence of it converges, and we have already dealt with the case of subsequences where the $(x_{n_k})$ or $(y_{n_k})$ converges. So we can safely assume , by renaming the index, that $x_n$ and $y_n$ do not have subsequences that converge to zero (in fact, to any number). A fortiori there exists $\delta > 0$ such that $\min(x_n^2, y_n^2) > \delta$ and the result follows. Q.E.D.
Bonus. We can also prove it with Taylor's theorem. Since $y_n^3 = x_n^3 + h_n$ (with $h_n \to 0$) and we can assume $|x_n| >  \delta > 0$ for all $n,$ we see that
$$
y_n = x_n + \dfrac{1}{3(c_n)^{\frac{2}{3}}} h_n
$$
with $c_n$ lying between $x_n^3$ and $x_n^3 + h_n,$ since $h_n$ converges to zero, we can assume $n$ large enough so that $|c_n| > \dfrac{\delta}{2}.$ In any case, we wrote $y_n = x_n + e_n$ with $e_n \to 0$ and we are done.
A: Suppose $x_n^3 - y_n^3\to 0$ and $f:\mathbb R\to \mathbb R$ is uniformly continuous. Then $f(x_n^3) - f(y_n^3)\to 0.$ So we're done if we can show $f(x)=x^{1/3}$ is uniformly continuous on $\mathbb R.$
To do this, note that $f$ is continuous on $\mathbb R,$ hence it is uniformly continuous on $[-1,1].$ Also $f$ is uniformly continuous on $[1,\infty).$  This follows from the MVT because, as you can verify, $|f'| \le 1/3$ on $[1,\infty).$ The same estimate holds on $(-\infty,-1].$ It follows that $f$ is uniformly continuous on $(-\infty,-1]\cup[-1,1]\cup [1,\infty) =\mathbb R,$ and we're done.
A: Suppose that you don't have $\lim_{n\to\infty}x_n-y_n=0$. Then there is some $r>0$ such that $|x_n-y_n|\geqslant r$ for infinitely many $n$'s. For each such $n$, you have\begin{align}x_n^{\,2}+x_ny_n+y_n^{\,2}&=\frac34(x_n+y_n)^2+\frac14(x_n-y_n)^2\\&\geqslant\frac{r^2}4\end{align}and therefore\begin{align}\left|x_n^{\,3}-y_n^{\,3}\right|&=\left(x_n^{\,2}+x_ny_n+y_n^{\,2}\right)|x_n-y_n|\\&\geqslant\frac{r^2}4\times r\\&=\frac{r^3}4.\end{align}So, you also don't have $\lim_{n\to\infty}x_n^{\,3}-y_n^{\,3}=0$.
A: For all $a,b \in \mathbb{R}$ ,$a^2+b^2 +ab \ge \frac{1}{4}(a-b)^2$, hence $$|a^3-b^3| =|a-b||a^2+ab+b^2| \ge \frac{1}{4}|a-b|^3$$
Besides $ x_n^3-y_n^3 \rightarrow 0$ means that $ \lim |x_n^3-y_n^3| =0$, so by squeeze theorem, $\lim \frac{1}{4}|x_n-y_n|^3=0$, or
$$\lim_{n \rightarrow +\infty} |x_n-y_n|=0 $$
