Another of $\frac{1^2}{1^2}+\frac{1-2^2+3^2-4^2}{1+2^2+3^2+4^2}+\cdots=\frac{\pi}{4}$ type expressable in cube? Gregory and Leibniz formula
(1)
$$-\sum_{m=1}^{\infty}\frac{(-1)^m}{2m-1}=\frac{\pi}{4}$$
We found another series equivalent to (1)
This is expressed in term of square numbers
$$-\sum_{m}^{\infty}\frac{\sum_{n=1}^{3m-2}(-1)^nn^2
}{\sum_{n=1}^{3m-2}(+1)^nn^2}=\frac{\pi}{4}$$
$$\frac{1^2}{1^2}+\frac{1-2^2+3^2-4^2}{1+2^2+3^2+4^2}+\frac{1-2^2+3^2-4^2+5^2-6^2+7^2}{1^2+2^2+3^2+4^2+5^2+6^2+7^2}+\cdots=\frac{\pi}{4}$$
Is there another series equivalent to (1) but expressable in term of cube numbers?

A proof of mahdi's result:
$$ \sum_{k=1}^{n}k^3 = \frac{n^2(n+1)^2}{4}\tag{1}$$
$$ \sum_{k=1}^{2n}(-1)^{k+1} k^3 = -n^2(4n+3),\qquad \sum_{k=1}^{2n-1}(-1)^{k+1} k^3 = n^2(4n-3)\tag{2}$$
lead to:
$$\begin{eqnarray*}\sum_{n\geq 1}\frac{1^3-2^3+\ldots}{1^3+2^3+\ldots+n^3} &=& \sum_{m\geq 1}\frac{-(4m+3)}{(2m+1)^2}+\sum_{m\geq 1}\frac{(4m-3)}{(2m-1)^2}\\&=&3+\sum_{m\geq 1}\frac{(4m-3)-(4m-1)}{(2n-1)^2}\\&=&3-2\sum_{m\geq 1}\frac{1}{(2n-1)^2}=\color{red}{3-\frac{\pi^2}{4}}.\tag{3}\end{eqnarray*}$$
