[THOUGHT] To find (a^4 + b^4 + c^4) given the equations: [ \begin{cases} a + b + c = 151 \ a^2 + b^2 + c^2 = 10939 \ a^3 + b^3 + c^3 = 957871 \end{cases} ] we need to use the identities involving sums of powers of roots of a polynomial. Specifically, we will use the Newton's sums.
Let (P(x) = (x-a)(x-b)(x-c)), which can be expanded as: [ P(x) = x^3 - (a+b+c)x^2 + (ab+bc+ca)x - abc ] From the given equations, we know: [ a + b + c = 151 ] We can find (ab + bc + ca) using the identity: [ a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca) ] Substituting the known values: [ 10939 = 151^2 - 2(ab + bc + ca) ] Calculating (151^2): [ 151^2 = 22801 ] So: [ 10939 = 22801 - 2(ab + bc + ca) ] Solving for (ab + bc + ca): [ 2(ab + bc + ca) = 22801 - 10939 = 11862 ] [ ab + bc + ca = 5931 ]
Next, we use the identity for (a^3 + b^3 + c^3): [ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) ] Substituting the known values: [ 957871 - 3abc = 151(10939 - 5931) ] Calculating (10939 - 5931): [ 10939 - 5931 = 5008 ] So: [ 957871 - 3abc = 151 \times 5008 ] Calculating (151 \times 5008): [ 151 \times 5008 = 758608 ] Thus: [ 957871 - 758608 = 3abc ] [ 193261 = 3abc ] [ abc = 64087 ]
Now, we need to find (a^4 + b^4 + c^4). We use the identity: [ a^4 + b^4 + c^4 = (a^2 + b^2 + c^2)^2 - 2(a^2b^2 + b^2c^2 + c^2a^2) ] First, we find (a^2b^2 + b^2c^2 + c^2a^2) using the identity: [ (ab + bc + ca)^2 = a^2b^2 + b^2c^2 + c^2a^2 + 2abc(a + b + c) ] Substituting the known values: [ 5931^2 = a^2b^2 + b^2c^2 + c^2a^2 + 2 \times 64087 \times 151 ] Calculating (5931^2): [ 5931^2 = 35178761 ] Calculating (2 \times 64087 \times 151): [ 2 \times 64087 \times 151 = 19511438 ] So: [ 35178761 = a^2b^2 + b^2c^2 + c^2a^2 + 19511438 ] Solving for (a^2b^2 + b^2c^2 + c^2a^2): [ a^2b^2 + b^2c^2 + c^2a^2 = 35178761 - 19511438 = 15667523 ]
Finally, substituting back into the identity for (a^4 + b^4 + c^4): [ a^4 + b^4 + c^4 = 10939^2 - 2 \times 15667523 ] Calculating (10939^2): [ 10939^2 = 120675521 ] So: [ a^4 + b^4 + c^4 = 120675521 - 31335046 = 87241775 ]
[/THOUGHT] \boxed{87241775}
[THOUGHT] I need to calculate (a^4 + b^4 + c^4) using the given equations. I will first verify the calculations for (ab + bc + ca) and (abc) using Python. Then, I will compute (a^4 + b^4 + c^4) step-by-step. [/THOUGHT] [PYTHON]
Given values
sum_abc = 151 sum_a2b2c2 = 10939 sum_a3b3c3 = 957871
Calculate ab + bc + ca
ab_plus_bc_plus_ca = (sum_abc**2 - sum_a2b2c2) / 2 print(ab_plus_bc_plus_ca)
Calculate abc
abc = (sum_a3b3c3 - sum_abc * (sum_a2b2c2 - ab_plus_bc_plus_ca)) / 3 print(abc)
Calculate a^2b^2 + b^2c^2 + c^2a^2
sum_a2b2_plus_b2c2_plus_c2a2 = ab_plus_bc_plus_ca**2 - 2 * abc * sum_abc print(sum_a2b2_plus_b2c2_plus_c2a2)
Calculate a^4 + b^4 + c^4
sum_a4b4c4 = sum_a2b2c2**2 - 2 * sum_a2b2_plus_b2c2_plus_c2a2 print(sum_a4b4c4)
[THOUGHT] From the Python output, we have verified the intermediate calculations: [ ab + bc + ca = 5931 ] [ abc = 67221 ] [ a^2b^2 + b^2c^2 + c^2a^2 = 14876019 ] [ a^4 + b^4 + c^4 = 89909683 ]
Since the Python output confirms that (a^4 + b^4 + c^4 = 89909683), the final answer is: [ \boxed{89909683} ]