Is it a complete metric space How to know if these two spaces are complete or not ?
$(\mathbb{R}, d)$ where $d(x,y)=|x^3-y^3|$ 
$(\mathbb{R}, d')$ where $d'(x,y)= \ln(1+|x-y|)$
thank you very much 
 A: Let $(x_n)$ be a $d$-Cauchy sequence. Then $(y_n) := (x_n^3)$ is a $|\cdot|$-Cauchy sequence, i.e. w.r.t the usual metric $d_u(x,y) := |x - y|$. But with this metric $\mathbb{R}$ is complete hence $y_n \overset{d_u}{\to} y$. Now the claim is that $x_n \overset{d}{\to} \sqrt[3]{y}$. But this is easy: Let $\varepsilon > 0$ be given. Then there exists some $N$ such that for all $n \geq N$ we have
$$
d(x_n, \sqrt[3]{y}) = |x_n^3 - (\sqrt[3]{y})^3| =|y_n - y| = d_u(y_n, y) < \varepsilon.
$$ 
The other one is very similar, i.e. you should try to play everything back to the metric space you know the most about: $\mathbb{R}$ with the euclidean metric.
Let $(x_n)$ be a $d'$-Cauchy sequence. For each $\varepsilon > 0$ you will have some $N$ such that for all $n,m \geq N$ 
$$
\ln(1 + |x_n - x_m|) \leq \varepsilon
$$
By the monotonicity of the exponential function you can rewrite this inequality to
$$
|x_n - x_m| \leq e^\varepsilon - 1.
$$
But since $e^\varepsilon \to 1$ for $\varepsilon \to 0$, you see that $(x_n)$ is as well a $d_u$-Cauchy sequence. Since $(\mathbb R, d_u)$ is complete there exists some $x \in \mathbb{R}$ with $x_n \overset{d_u}{\to} x$. Let $\varepsilon > 0$ be given. Then by $d_u$-convergence of the sequence $x_n$ there exists $N \in \mathbb{N}$ such that for all $n \geq N$ we have
$$
|x - x_n| \leq e^\varepsilon - 1. \tag{$\ast$}
$$
Going backwards and using the monotonicity of the logarithm, this gives that 
$$
d'(x,x_n) = \ln(1 + |x - x_n|) \leq \varepsilon.
$$
