Combining the identity (the composition of the polynomial $P_3(x)=\frac{x(x+1)(2x+1)}{6}$ with the arithmetical function $N(n)=n$ and the known proof by induction)
$$1^2+2^2+\ldots+n^2=\frac{n(n+1)(2n+1)}{6}$$
with the substitution $n=\frac{\sigma(n)}{2}$, where $\sigma(n)=\sum_{d\mid n}d$ is the sum of divisors function, thus is the condition to be perfect we obtain from
$$\frac{n(n+1)(2n+1)}{6}=1^2+2^2+\ldots+n^2=\frac{\frac{\sigma(n)}{2}\left( \frac{\sigma(n)}{2}+1\right) (\sigma(n)+1)}{6}$$ the following
Proposition 2. If $n$ is a perfect number then $$4n(n+1)(2n+1)=\sigma(n)(1+\sigma(n))(2+\sigma(n)).$$
I believe that my question is only about algebra, not about perfect numbers, since is the resolution of a cubic equation.
Question. Can you show that if an integer $m\geq 1$ satisfies $$\sigma(m)^3+3\sigma(m)^2+2\sigma(m)-4m(m+1)(2m+1)=0$$ then $m$ is perfect?
I know that I need to use Cardano's identities, first I need take $\sigma(n)=y(n)-1$
Remark (it isn't neccesary read to solve previous question). Using the corresponding (similar polynomials $P_0(x)$, $P_1(x)$ and $P_4(x)$) the known power finite series and the definition/substitution of the condition to be a perfect number I know how prove the corresponding Proposition 0, Proposition 1 and Proposition 2 as equivalences if and only if with perfect numbers. In principle , there is nothing special in these polynomials and identities (you can use for example $1+7n+n^2=1+7n+n^2$, and also for other sequences with form as prime numbers $\sigma(n)=n+1$ or almost-perfect numbers), and I believe that don't provide us additional information since the substitution is obvious when we divide (using Euler-Fermat), since the satements that we can prove seem the obvious congruence $0\equiv 0$. From Galois theory it is know that isn't possible solve by radicals a polynomial equation with degree $\geq 5$, and I don't know what happens if I consider polynomials $P(n)\in \mathbb{C}[x]$, this is with coefficients of the form $a+bi$, which $a,b$ are integers. Thus in principle, the condition to be perfect in my question seems accessory, and I believe that is only an exercise of basic algebra: how solve the tedious computations of a cubic equation.