a + b + c = 4 a^2 + b^2 + c^2 = 10 a^3…

✅ Answer

The value of \(a^4 + b^4 + c^4\) is 50.

🔢 Step-by-Step Solution

1. Given the power sums:

   p1 = a + b + c = 4
   p2 = a^2 + b^2 + c^2 = 10
   p3 = a^3 + b^3 + c^3 = 22
   

1. Use Newton’s identity for p2 to find s2 = ab+bc+ca:

   p2 = s1^2 - 2 s2
   => 10 = 4^2 - 2 s2
   => 10 = 16 - 2 s2
   => 2 s2 = 6
   => s2 = 3
   

1. Use Newton’s identity for p3 to find s3 = abc:

   p3 = s1 p2 - s2 p1 + 3 s3
   => 22 = 4*10 - 3*4 + 3 s3
   => 22 = 40 - 12 + 3 s3
   => 22 = 28 + 3 s3
   => 3 s3 = -6
   => s3 = -2
   

1. For p4 (with only three variables, the identity is p4 - s1 p3 + s2 p2 - s3 p1 = 0), compute:

   p4 = s1 p3 - s2 p2 + s3 p1
   => p4 = 4*22 - 3*10 + (-2)*4
   => p4 = 88 - 30 - 8
   => p4 = 50
   

💡 Explanation

We used Newton’s identities, which relate the power sums \(p_k = a^k+b^k+c^k\) to the elementary symmetric sums \(s_1=a+b+c\), \(s_2=ab+bc+ca\), \(s_3=abc\). For three variables, these identities let us solve for the symmetric sums from the given power sums, then compute higher power sums (here \(p_4\)) directly from those symmetric sums.

📌 Key Points

• Given p1 = 4, p2 = 10, p3 = 22.
• Found s2 = 3 and s3 = -2 using Newton’s identities.
• Final result: a^4 + b^4 + c^4 = 50.