Math Problem Statement

Given the system of equations: a + b + c = 4, a^2 + b^2 + c^2 = 10, a^3 + b^3 + c^3 = 22, find a^4 + b^4 + c^4.

Solution

We are given the following system of equations:

a+b+c=4a + b + c = 4 a2+b2+c2=10a^2 + b^2 + c^2 = 10 a3+b3+c3=22a^3 + b^3 + c^3 = 22

Our goal is to determine the value of:

a4+b4+c4=?a^4 + b^4 + c^4 = ?

This problem can be approached systematically using Newton’s sums, which relate the symmetric sums of roots with the powers of roots. Let's go step-by-step to solve for a4+b4+c4a^4 + b^4 + c^4.


Step-by-Step Solution using Newton's Sums

We define the symmetric sums:

  • P1=a+b+c=4P_1 = a + b + c = 4,
  • P2=a2+b2+c2=10P_2 = a^2 + b^2 + c^2 = 10,
  • P3=a3+b3+c3=22P_3 = a^3 + b^3 + c^3 = 22,
  • We want to find P4=a4+b4+c4P_4 = a^4 + b^4 + c^4.

Newton's sums formula for higher-order terms:

Pn=(a+b+c)Pn1k=1n2PkSnk+nSnP_n = (a + b + c) P_{n-1} - \sum_{k=1}^{n-2} P_k \cdot S_{n-k} + n \cdot S_n

Since we are dealing with three terms a,b,ca, b, c, the coefficients SkS_k for k>1k > 1 are generally related to sums of products like ab+bc+caab + bc + ca.

Step 1: Use identity for squares.

We have the identity:

a2+b2+c2=(a+b+c)22(ab+bc+ca)a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca)

Plugging in known values:

10=422(ab+bc+ca)10 = 4^2 - 2(ab + bc + ca)

10=162(ab+bc+ca)10 = 16 - 2(ab + bc + ca)

2(ab+bc+ca)=6ab+bc+ca=32(ab + bc + ca) = 6 \quad \Rightarrow \quad ab + bc + ca = 3

Step 2: Express higher powers.

Now, using the Newton's sum formula for P3P_3, we get:

P3=(a+b+c)P2(ab+bc+ca)P1P_3 = (a + b + c) P_2 - (ab + bc + ca) P_1

Plugging in values:

22=4103422 = 4 \cdot 10 - 3 \cdot 4

22=401222 = 40 - 12

This is consistent. Now let’s use the same formula to find P4P_4.

Step 3: Find P4P_4.

Using the Newton's sum formula for P4P_4:

P4=(a+b+c)P3(ab+bc+ca)P2P_4 = (a + b + c) P_3 - (ab + bc + ca) P_2

Substitute the known values:

P4=422310P_4 = 4 \cdot 22 - 3 \cdot 10

P4=8830=58P_4 = 88 - 30 = 58


Final Answer

a4+b4+c4=58a^4 + b^4 + c^4 = 58


Do you need further explanation, or would you like to explore another method?
Here are some related questions for practice:

  1. How do you derive Newton's sums for polynomial roots in general?
  2. What are the identities for a5+b5+c5a^5 + b^5 + c^5 using the same approach?
  3. How is the symmetric sum ab+bc+caab + bc + ca derived from roots of a cubic equation?
  4. What are the limits of Newton’s sums? Can they be applied to non-polynomial roots?
  5. How would you solve this system if you knew the explicit roots a,b,a, b, and cc?

Tip: Newton’s sums are very useful for symmetric polynomial problems. Mastering them helps solve high-degree equations efficiently.

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Math Problem Analysis

Mathematical Concepts

Algebra
Symmetric Polynomials
Newton's Sums

Formulas

P_n = (a + b + c) P_{n-1} - ∑ P_k * S_{n-k} + n * S_n
a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca)

Theorems

Newton's Sums

Suitable Grade Level

Grades 10-12