Sorry for the late reply, I was inactive for about 4 years. Anyway, most of Ramanujan's results are special cases of more general and perhaps deeper phenomena.
I. 3rd powers
For your question, a more general algebraic identity is,
$$A^3+B^3+C^3+D^3 = (a^3+b^3+c^3+d^3)(x^2+wy^2)^3$$
and $(A,B,C,D)$ are quadratic forms,
\begin{align}
A &= ax^2-v_1xy+bwy^2\\
B &= bx^2+v_1xy+awy^2\\
C &= cx^2+v_2xy+dwy^2\\
D &= dx^2-v_2xy+cwy^2
\end{align}
where {$v_1,\, v_2,\, w$} = {$c^2-d^2,\; a^2-b^2,\; (a+b)(c+d).$} I gave this identity to Mathworld about 20 years ago.
Thus, given a single example of $a^3+b^3+c^3+d^3 =N$ where $N=0$, a cube, or any number of cubes, then one can generate an infinite more via the LHS. For example, try it with the well-known $3^3+4^3+5^3+(-6)^3 = 0$, permute these variables $(a,b,c,d)$ in any order you wish, and you'll have your own Ramanujan-type identity of sums of cubes.
II. 4th powers
Ramanujan also gave,
$$(2x^2+12xy-6y^2)^4+(2x^2-12xy-6y^2)^4+(4x^2-12y^2)^4+(4x^2+12y^2)^4+(3x^2+9y^2)^4 = (5x^2+15y^2)^4$$
that for $(x,y)=(1,0)$ yields the nice,
$$2^4+2^4+4^4+4^4+3^4 = 5^4$$
The general identity (which I also gave to Mathworld) is,
$$A^4+B^4+C^4 = (a^4+b^4+c^4)(x^2+3y^2)^4$$
and $(A,B,C)$ are also quadratic forms,
\begin{align}
A &= ax^2+2(b+c)xy-3ay^2\\
B &= bx^2-2(a+c)xy-3by^2\\
C &= cx^2-2(a-b)xy-3cy^2\\
\end{align}
where $\color{brown}{a+b = c}$. Thus, given a single $a^4+b^4+c^4+d^4+e^4 =f^4$ where $a+b = c$, one can find an infinite more.
P.S. There's actually another cubic identity (as well as a $5$th-deg, but I'm too sleepy. I'll add it soon.) More details can be found at "Generalizing Ramanujan's sum of cubes identity?"