Question

a + b + c = 4
a² + b² + c² = 10
a³ + b³ + c³ = 22
a⁴ + b⁴ + c⁴ = ?

Explanation:

Step1: Recall the identity for square of sum

We know that \((a + b + c)^2=a^2 + b^2 + c^2+2(ab + bc + ca)\). We are given \(a + b + c = 4\), \(a^2 + b^2 + c^2=10\). Substitute these values into the identity:
\(4^2=10 + 2(ab + bc + ca)\)
\(16=10 + 2(ab + bc + ca)\)
Subtract 10 from both sides: \(16 - 10=2(ab + bc + ca)\)
\(6 = 2(ab + bc + ca)\)
Divide both sides by 2: \(ab + bc + ca=\frac{6}{2}=3\)

Step2: Recall the identity for sum of cubes

We know that \(a^3 + b^3 + c^3-3abc=(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)\). We are given \(a^3 + b^3 + c^3 = 22\), \(a + b + c = 4\), \(a^2 + b^2 + c^2=10\), \(ab + bc + ca = 3\). Substitute these values:
\(22-3abc=4\times(10 - 3)\)
\(22-3abc=4\times7\)
\(22-3abc = 28\)
Subtract 22 from both sides: \(- 3abc=28 - 22=6\)
Divide both sides by - 3: \(abc=\frac{6}{-3}=- 2\)

Step3: Find \(a^4 + b^4 + c^4\) using identity

We know that \(a^4 + b^4 + c^4=(a^2 + b^2 + c^2)^2-2(a^2b^2 + b^2c^2 + c^2a^2)\). First, we need to find \(a^2b^2 + b^2c^2 + c^2a^2\). We know that \((ab + bc + ca)^2=a^2b^2 + b^2c^2 + c^2a^2+2abc(a + b + c)\). We have \(ab + bc + ca = 3\), \(abc=-2\), \(a + b + c = 4\). Substitute these values:
\(3^2=a^2b^2 + b^2c^2 + c^2a^2+2\times(-2)\times4\)
\(9=a^2b^2 + b^2c^2 + c^2a^2-16\)
Add 16 to both sides: \(a^2b^2 + b^2c^2 + c^2a^2=9 + 16 = 25\)

Now, substitute \(a^2 + b^2 + c^2 = 10\) and \(a^2b^2 + b^2c^2 + c^2a^2 = 25\) into the formula for \(a^4 + b^4 + c^4\):
\(a^4 + b^4 + c^4=10^2-2\times25\)
\(a^4 + b^4 + c^4 = 100 - 50=50\)

Answer:

\(50\)