[THOUGHT] To find (a^4 + b^4 + c^4) for natural numbers (a, b, c) satisfying the given equations, we use symmetric sums and Newton's identities. First, we compute the elementary symmetric sums (S_1 = a + b + c), (S_2 = ab + bc + ca), and (S_3 = abc). Using the given equations, we find (S_1 = 151), (S_2 = 5931), and (S_3 = 67221). The sum of the fourth powers can be derived using the identity: [ a^4 + b^4 + c^4 = (a^2 + b^2 + c^2)^2 - 2(a^2b^2 + a^2c^2 + b^2c^2) ] where (a^2b^2 + a^2c^2 + b^2c^2 = S_2^2 - 2S_1S_3). Substituting the known values, we compute the result. Additionally, verifying by solving the cubic equation (x^3 - S_1x^2 + S_2x - S_3 = 0) confirms the roots (a = 21), (b = 33), and (c = 97). Calculating their fourth powers directly confirms the result.

[PYTHON]

Calculate the sum of fourth powers using symmetric sums

S1 = 151 S2 = 5931 S3 = 67221 sum_squares = 10939

Calculate a^2b^2 + a^2c^2 + b^2c^2

sum_sq_products = S2**2 - 2 * S1 * S3

Calculate sum of fourth powers

sum_fourth = sum_squares**2 - 2 * sum_sq_products print(sum_fourth)

Verify by computing individual fourth powers of roots 21, 33, 97

a, b, c = 21, 33, 97 verification = a4 + b4 + c**4 print(verification) [/PYTHON]

[THOUGHT] The Python output confirms that both methods yield the same result: (89909683). This value is consistent with the symmetric sum derivation and the direct computation of the fourth powers of the roots (21), (33), and (97). Therefore, the final answer is (\boxed{89909683}).

\boxed{89909683}