show that $$\sum_{t|n}(d(t))^3=\left(\sum_{t|n}d(t)\right)^2$$
where $d(n)$ is the number of positive divisors of $n$.
see this have simaler $$1^3+2^3+\cdots+n^3=\left(1+2+\cdots+n\right)^2$$
maybe this have something to do?
and How prove this nice equation.Thank you
If $n = p^k$ is a prime power, we have
$$\sum_{t\mid n} d(t)^3 = \sum_{j=0}^k (j+1)^3 = \left(\sum_{j=0}^k (j+1)\right)^2 = \left(\sum_{t\mid n} d(t)\right)^2$$
by the "similar" identity. For general $n$, use the multiplicativity of $d(\cdot)^m$, that is, for coprime $a,b$ we have $d(ab) = d(a)d(b)$, which becomes
$$\sum_{t\mid n} d(t)^3 = \prod_{p\mid n} \sum_{j= 0}^{v_p(n)} (j+1)^3 = \prod_{p\mid n} \left(\sum_{j=0}^{v_p(n)} (j+1)\right)^2 = \left(\sum_{t\mid n} d(t)\right)^2$$
since for a multiplicative function $f$, also the sum $F(n) = \sum_{t\mid n} f(t)$ is multiplicative.
