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Mathworld gives two parametrizations of Ramanujan : if $a+b+c=0$ then

$$a^4(b-c)^4+ b^4(c-a)^4+ c^4(a-b)^4= 2(ab+bc+ca)^4$$

and

$$(a^3+2abc)^4(b-c)^4+(b^3+2abc)^4(c-a)^4+(c^3+2abc)^4(a-b)^4=2(ab+ac+bc)^8$$

(equations 144 and 146). See also Bhargava, S. (1992) On a family of Ramanujan's formulas for sums of fourth powers. Ganita, 43 (1-2). pp. 63-67. [abstract] [summary]

The Ferrari Identity gives

$$(a^2+2ac-2bc-b^2)^4+ (b^2-2ba-2ca-c^2)^4+ (c^2+2cb+2ab-a^2)^4 = 2(a^2+b^2+c^2-ab+bc+ca)^4$$

There is also Ford's Theorem: if $S_j=\sum_{i=j\,(\text{mod}\,3)}(-1)^i\binom ki a^{k-i}b^i$ then $$(S_0-S_1)^4+(S_1-S_2)^4+(S_2-S_0)^4=2(a^2+ab+b^2)^{2k}$$

Ramanujan gave two parametrizations: if $a+b+c=0$ then

$$a^4(b-c)^4+ b^4(c-a)^4+ c^4(a-b)^4= 2(ab+bc+ca)^4$$

and

$$(a^3+2abc)^4(b-c)^4+(b^3+2abc)^4(c-a)^4+(c^3+2abc)^4(a-b)^4=2(ab+ac+bc)^8$$

(listed in Mathworld, equations 144 and 146). See also Bhargava, S. (1992) On a family of Ramanujan's formulas for sums of fourth powers. Ganita, 43 (1-2). pp. 63-67. [abstract] [summary]

The F. Ferrari Identity (1909) gives

$$(a^2+2ac-2bc-b^2)^4+ (b^2-2ba-2ca-c^2)^4+ (c^2+2cb+2ab-a^2)^4 = 2(a^2+b^2+c^2-ab+bc+ca)^4$$

There is also K. Ford's Theorem: if $S_j=\sum_{i=j\,(\text{mod}\,3)}(-1)^i\binom ki a^{k-i}b^i$ then $$(S_0-S_1)^4+(S_1-S_2)^4+(S_2-S_0)^4=2(a^2+ab+b^2)^{2k}$$

For $k=2$ and $k=4$, the Ford parametrizations are the same as the two Ramanujan ones. For $k=2$, this is simple, and for $k=4$, $S_0 = a^4-4 a b^3$, $S_1 = b^4-4 a^3 b$, $S_2 = 6 a^2 b^2$, and

$$\begin{align} S_0 - S_1 = -(c^3+2abc)(a-b)\\ S_1 - S_2 = -(b^3+2abc)(c-a)\\ S_2 - S_0 = -(a^3+2abc)(b-c)\\ \end{align}$$

where $c=-a-b$.

Edit: A complete discussion of these identities, by S. Ramanujan, F. Ferrari and Kevin Ford, plus further identities by S. Bhargava, may be found on pages 96 to 101 of Bruce Berndt's Ramanujan's Notebooks, Part IV (1994), at e.g. http://www.plouffe.fr/simon/math/Ramanujan%27s%20Notebooks%20IV.pdf.

Mathworld gives two parametrizations of Ramanujan : if $a+b+c=0$ then

$$a^4(b-c)^4+ b^4(c-a)^4+ c^4(a-b)^4= 2(ab+bc+ca)^4$$

and

$$(a^3+2abc)^4(b-c)^4+(b^3+2abc)^4(c-a)^4+(c^3+2abc)^4(a-b)^4=2(ab+ac+bc)^8$$

(equations 144 and 146).

The Ferrari Identity gives

$$(a^2+2ac-2bc-b^2)^4+ (b^2-2ba-2ca-c^2)^4+ (c^2+2cb+2ab-a^2)^4 = 2(a^2+b^2+c^2-ab+bc+ca)^4$$

There is also Ford's Theorem: if $S_j=\sum_{i=j\,(\text{mod}\,3)}(-1)^i\binom ki a^{k-i}b^i$ then $$(S_0-S_1)^4+(S_1-S_2)^4+(S_2-S_0)^4=2(a^2+ab+b^2)^{2k}$$

Mathworld gives two parametrizations of Ramanujan : if $a+b+c=0$ then

$$a^4(b-c)^4+ b^4(c-a)^4+ c^4(a-b)^4= 2(ab+bc+ca)^4$$

and

$$(a^3+2abc)^4(b-c)^4+(b^3+2abc)^4(c-a)^4+(c^3+2abc)^4(a-b)^4=2(ab+ac+bc)^8$$

(equations 144 and 146). See also Bhargava, S. (1992) On a family of Ramanujan's formulas for sums of fourth powers. Ganita, 43 (1-2). pp. 63-67. [abstract] [summary]

The Ferrari Identity gives

$$(a^2+2ac-2bc-b^2)^4+ (b^2-2ba-2ca-c^2)^4+ (c^2+2cb+2ab-a^2)^4 = 2(a^2+b^2+c^2-ab+bc+ca)^4$$

There is also Ford's Theorem: if $S_j=\sum_{i=j\,(\text{mod}\,3)}(-1)^i\binom ki a^{k-i}b^i$ then $$(S_0-S_1)^4+(S_1-S_2)^4+(S_2-S_0)^4=2(a^2+ab+b^2)^{2k}$$

Mathworld gives two parametrizations of Ramanujan : if $a+b+c=0$ then

$$a^4(b-c)^4+ b^4(c-a)^4+ c^4(a-b)^4= 2(ab+bc+ca)^4$$

and

$$(a^3+2abc)^4(b-c)^4+(b^3+2abc)^4(c-a)^4+(c^3+2abc)^4(a-b)^4=2(ab+ac+bc)^8$$

(equations 144 and 146).

And the Ferrari Identity gives

$$(a^2+2ac-2bc-b^2)^4+ (b^2-2ba-2ca-c^2)^4+ (c^2+2cb+2ab-a^2)^4 = 2(a^2+b^2+c^2-ab+bc+ca)^4$$

There is also Ford's Theorem: if $$S_j=\sum_{i=j\,(\text{mod}\,3)}(-1)^i\binom ki a^{k-i}b^i$$
then $$(S_0-S_1)^4+(S_1-S_2)^4+(S_2-S_0)^4=2(a^2+ab+b^2)^{2k}$$

Mathworld gives two parametrizations of Ramanujan : if $a+b+c=0$ then

$$a^4(b-c)^4+ b^4(c-a)^4+ c^4(a-b)^4= 2(ab+bc+ca)^4$$

and

$$(a^3+2abc)^4(b-c)^4+(b^3+2abc)^4(c-a)^4+(c^3+2abc)^4(a-b)^4=2(ab+ac+bc)^8$$

(equations 144 and 146).

The Ferrari Identity gives

$$(a^2+2ac-2bc-b^2)^4+ (b^2-2ba-2ca-c^2)^4+ (c^2+2cb+2ab-a^2)^4 = 2(a^2+b^2+c^2-ab+bc+ca)^4$$

There is also Ford's Theorem: if $S_j=\sum_{i=j\,(\text{mod}\,3)}(-1)^i\binom ki a^{k-i}b^i$ then $$(S_0-S_1)^4+(S_1-S_2)^4+(S_2-S_0)^4=2(a^2+ab+b^2)^{2k}$$