[THOUGHT] To find ( a^4 + b^4 + c^4 ) given the equations:
- ( a + b + c = 151 )
- ( a^2 + b^2 + c^2 = 10939 )
- ( a^3 + b^3 + c^3 = 957871 )
I will use Newton's identities to relate these sums to the elementary symmetric polynomials. First, I need to find the elementary symmetric sums ( S_1 = a + b + c ), ( S_2 = ab + bc + ca ), and ( S_3 = abc ). Then, I can use these to find higher power sums like ( a^4 + b^4 + c^4 ).
From the given:
- ( S_1 = 151 )
- ( a^2 + b^2 + c^2 = S_1^2 - 2S_2 ), so ( 10939 = 151^2 - 2S_2 ). Solving for ( S_2 ): ( S_2 = (151^2 - 10939)/2 ).
Similarly, using the identity for ( a^3 + b^3 + c^3 ): ( a^3 + b^3 + c^3 = S_1^3 - 3S_1S_2 + 3S_3 ), so ( 957871 = 151^3 - 3 \cdot 151 \cdot S_2 + 3S_3 ). Solving for ( S_3 ).
Once I have ( S_1, S_2, S_3 ), I can use the recurrence relation for power sums to find ( a^4 + b^4 + c^4 ).
[/THOUGHT] [PYTHON] from sympy import symbols, solve, simplify
Given sums
S1 = 151 sum_sq = 10939 sum_cu = 957871
Calculate S2 = ab + bc + ca
S2 = (S1**2 - sum_sq) // 2
Calculate S3 = abc
S3 = (sum_cu - S1**3 + 3 * S1 * S2) // 3
Now, use the recurrence relation for power sums:
p_k = S1 * p_{k-1} - S2 * p_{k-2} + S3 * p_{k-3}
We know p1 = S1, p2 = sum_sq, p3 = sum_cu
p1 = S1 p2 = sum_sq p3 = sum_cu
Calculate p4 = a^4 + b^4 + c^4
p4 = S1 * p3 - S2 * p2 + S3 * p1
print(p4) [/PYTHON]
[THOUGHT] The Python output confirms that ( a^4 + b^4 + c^4 = 89909683 ). [/THOUGHT] \boxed{89909683}