Algebraic Analysis of Power Sums for Three Variables

Mathematical Problem Overview: Power Sums of Three Variables

  • This problem involves a system of three non-linear equations with three variables (aa, bb, and cc) and requires finding the value of a specific expression related to the fourth power of these variables, based on the patterns established by the first three equations.
  • The given information is as follows:
    • Sum of the variables: a+b+c=4a + b + c = 4
    • Sum of the squares: a2+b2+c2=10a^2 + b^2 + c^2 = 10
    • Sum of the cubes: a3+b3+c3=22a^3 + b^3 + c^3 = 22
  • The objective is to determine the value of the fourth power sum: a4+b4+c4=?a^4 + b^4 + c^4 = ?

Theoretical Framework: Newton-Girard Formulas

  • The most efficient method to solve for higher power sums without solving for individual variables (aa, bb, and cc) is using the Newton-Girard Formulas.

  • These formulas relate power sums (pnp_n) to elementary symmetric polynomials (ene_n).

  • Let:

    • p1=a+b+cp_1 = a + b + c
    • p2=a2+b2+c2p_2 = a^2 + b^2 + c^2
    • p3=a3+b3+c3p_3 = a^3 + b^3 + c^3
    • pk=ak+bk+ckp_k = a^k + b^k + c^k
  • Let the elementary symmetric polynomials be:

    • e1=a+b+ce_1 = a + b + c
    • e2=ab+bc+cae_2 = ab + bc + ca
    • e3=abce_3 = abc
  • The recursive Newton-Girard relations for three variables are defined by:

    • p1e1=0p_1 - e_1 = 0
    • p2e1p1+2e2=0p_2 - e_1p_1 + 2e_2 = 0
    • p3e1p2+e2p13e3=0p_3 - e_1p_2 + e_2p_1 - 3e_3 = 0
    • pne1pn1+e2pn2e3pn3=0p_n - e_1p_{n-1} + e_2p_{n-2} - e_3p_{n-3} = 0 (for n3n \ge 3)

Step-by-Step Calculation of Elementary Symmetric Polynomials

  • Step 1: Determine e1e_1

    • From the first equation and the definition of e1e_1:
    • e1=a+b+c=4e_1 = a + b + c = 4
  • Step 2: Determine e2e_2

    • Using the relationship p2=e1p12e2p_2 = e_1p_1 - 2e_2:
    • Substitute the known values: 10=(4)(4)2e210 = (4)(4) - 2e_2
    • Simplify the equation: 10=162e210 = 16 - 2e_2
    • Subtract 16 from both sides: 6=2e2-6 = -2e_2
    • Divide by -2: e2=3e_2 = 3
  • Step 3: Determine e3e_3

    • Using the relationship p3=e1p2e2p1+3e3p_3 = e_1p_2 - e_2p_1 + 3e_3:
    • Substitute the known values: 22=(4)(10)(3)(4)+3e322 = (4)(10) - (3)(4) + 3e_3
    • Simplify the equation: 22=4012+3e322 = 40 - 12 + 3e_3
    • Further simplification: 22=28+3e322 = 28 + 3e_3
    • Subtract 28 from both sides: 6=3e3-6 = 3e_3
    • Divide by 3: e3=2e_3 = -2

Determination of the Fourth Power Sum

  • To find p4=a4+b4+c4p_4 = a^4 + b^4 + c^4, we utilize the recursive formula for n=4n = 4:
    • p4=e1p3e2p2+e3p1p_4 = e_1p_3 - e_2p_2 + e_3p_1
  • Substitute the calculated values for e1e_1, e2e_2, and e3e_3, along with the given values for p1p_1, p2p_2, and p3p_3:
    • p4=(4)(22)(3)(10)+(2)(4)p_4 = (4)(22) - (3)(10) + (-2)(4)
  • Perform the multiplications:
    • (4)(22)=88(4)(22) = 88
    • (3)(10)=30(3)(10) = 30
    • (2)(4)=8(-2)(4) = -8
  • Calculate the final result:
    • p4=88308p_4 = 88 - 30 - 8
    • p4=588p_4 = 58 - 8
    • p4=50p_4 = 50

Summary of Results

  • Based on the algebraic properties of symmetric polynomials:
    • The sum of the variables is 44.
    • The sum of their pairwise products is 33.
    • The product of the three variables is 2-2.
    • The value of the expression a4+b4+c4a^4 + b^4 + c^4 is exactly 5050.