Solving Systems of Power Sum Equations for Three Variables

Problem Statement and Identified Goal

  • The transcript identifies a system of three nonlinear equations containing three variables: aa, bb, and cc.
  • The primary objective is to find the value of a certain expression based on the given powers of the variables.
  • Given Equations:
    • Equation 1: a+b+c=4a + b + c = 4
    • Equation 2: a2+b2+c2=10a^2 + b^2 + c^2 = 10
    • Equation 3: a3+b3+c3=22a^3 + b^3 + c^3 = 22
  • Target Expression: The transcript lists "a² + b + c 4 =?". Given the progression of powers in the previous lines (1st, 2nd, and 3rd powers), this is interpreted as a request for the sum of the fourth powers: a4+b4+c4a^4 + b^4 + c^4.

Elementary Symmetric Polynomials of the Variables aa, bb, and cc

  • To solve for higher power sums, we utilize the elementary symmetric polynomials for three variables, defined as:
    • e1=a+b+ce_1 = a + b + c
    • e2=ab+bc+cae_2 = ab + bc + ca
    • e3=abce_3 = abc
  • From the first given equation, we immediately establish:
    • e1=4e_1 = 4
  • The higher-order polynomials e2e_2 and e3e_3 must be derived from the second and third power sums provided in the transcript.

Newton's Identities and Recurrence Relations

  • Newton's Identities provide a relationship between power sums pk=ak+bk+ckp_k = a^k + b^k + c^k and elementary symmetric polynomials eke_k.
  • The recursive formulas for three variables are as follows:
    • p1e1=0p_1 - e_1 = 0
    • p2e1p1+2e2=0p_2 - e_1 p_1 + 2e_2 = 0
    • p3e1p2+e2p13e3=0p_3 - e_1 p_2 + e_2 p_1 - 3e_3 = 0
    • p4e1p3+e2p2e3p1+4e4=0p_4 - e_1 p_3 + e_2 p_2 - e_3 p_1 + 4e_4 = 0
  • Note: For a system of three variables, ek=0e_k = 0 for all k>3k > 3. Therefore, e4=0e_4 = 0, simplifying the equation for p4p_4.

Step-by-Step Derivation of Symmetric Polynomial Values

  • Calculation of e2e_2:
    • Using the identity associated with p2p_2:
    • 10(4×4)+2e2=010 - (4 \times 4) + 2e_2 = 0
    • 1016+2e2=010 - 16 + 2e_2 = 0
    • 6+2e2=0-6 + 2e_2 = 0
    • 2e2=62e_2 = 6
    • e2=3e_2 = 3
  • Verification of e2e_2 using standard expansion:
    • (a+b+c)2=a2+b2+c2+2(ab+bc+ca)(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)
    • 42=10+2e24^2 = 10 + 2e_2
    • 16=10+2e26=2e2e2=316 = 10 + 2e_2 \rightarrow 6 = 2e_2 \rightarrow e_2 = 3
  • Calculation of e3e_3:
    • Using the identity associated with p3p_3:
    • 22(4×10)+(3×4)3e3=022 - (4 \times 10) + (3 \times 4) - 3e_3 = 0
    • 2240+123e3=022 - 40 + 12 - 3e_3 = 0
    • 18+123e3=0-18 + 12 - 3e_3 = 0
    • 63e3=0-6 - 3e_3 = 0
    • 3e3=63e_3 = -6
    • e3=2e_3 = -2
  • Verification of e3e_3 using the cubic sum formula:
    • a3+b3+c33abc=(a+b+c)(a2+b2+c2(ab+bc+ca))a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - (ab + bc + ca))
    • 223e3=(4)(103)22 - 3e_3 = (4)(10 - 3)
    • 223e3=4×722 - 3e_3 = 4 \times 7
    • 223e3=2822 - 3e_3 = 28
    • 3e3=6e3=2-3e_3 = 6 \rightarrow e_3 = -2

Calculation of the Fourth Power Sum a4+b4+c4a^4 + b^4 + c^4

  • With e1=4e_1 = 4, e2=3e_2 = 3, and e3=2e_3 = -2, we can now solve for p4=a4+b4+c4p_4 = a^4 + b^4 + c^4 using the fourth Newton Identity where e4=0e_4 = 0:
    • p4e1p3+e2p2e3p1=0p_4 - e_1 p_3 + e_2 p_2 - e_3 p_1 = 0
    • Substitute the known values into the equation:
    • p4(4×22)+(3×10)(2×4)=0p_4 - (4 \times 22) + (3 \times 10) - (-2 \times 4) = 0
  • Perform the arithmetic operations:
    • p488+30+8=0p_4 - 88 + 30 + 8 = 0
    • p458+8=0p_4 - 58 + 8 = 0
    • p450=0p_4 - 50 = 0
  • Solve for p4p_4:
    • p4=50p_4 = 50

Summary of Calculated Results

  • The elementary symmetric polynomials for the set {a,b,c}\{a, b, c\} are:
    • Sum (e1e_1): 44
    • Sum of pairwise products (e2e_2): 33
    • Product of all variables (e3e_3): 2-2
  • These values correspond to the coefficients of the characteristic cubic equation x34x2+3x+2=0x^3 - 4x^2 + 3x + 2 = 0, whose roots are the values of aa, bb, and cc.
  • The value of the fourth power sum is determined to be:
    • a4+b4+c4=50a^4 + b^4 + c^4 = 50