Complete measure spaces and convergence in $L^p(\mu)$

Question

Problem: Let $(X, \mathcal{M}, \mu)$ be a complete measure space. Prove that, if $f_n(x)$ converges to $f(x)$ in $L^3(\mu)$, then $f^3_n(x)$ converges to $f^3(x)$ in $L^1(\mu)$.

My attempt: Let $\|f\|_p = \left(\int_X|f|^p\,d\mu\right)^{1/p}$. Then $$f \in L^3(\mu) \implies \|f\|_3 < \infty \implies \|f^3\|_1 = \|f\|^3_3 < \infty \implies f^3 \in L^1(\mu).$$

My issue: I don't feel like this proof is complete or correct; I didn't use the fact that $(X, \mathcal{M}, \mu)$ is a complete measure space (at least, not explicitly; if I implicitly assumed it, I'm unsure where), and I'm not confident that my first and last implications are correct (I'm used to seeing things like $L^3[0, 1]$, but not $L^3(\mu)$).

My request: If anyone could guide my apparent misunderstandings of the concepts used here, I would greatly appreciate it. If I'm on the wrong track, I'd also appreciate a nudge in the right direction (but NOT a full solution).

Extra info: I'm using Royden's Real Analysis

Answer 1

As suchan mentioned in the comment section, you need to show that $\|f_n^3 - f^3\|_{L^1} \to 0$.

Hint1: $$|f_n^3 - f^3| = |f_n - f||f_n^2 + f^2 +f_nf|.$$

Hint2:

Use Holder's inequality with $p = 3$ and $q = \frac{3}{2}$.