Question

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

Explanation:

Step1: Find \( ab + bc + ca \)

We know the identity \( (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) \). Substitute \( a + b + c = 4 \), \( a^2 + b^2 + c^2 = 10 \) into it:
\( 4^2 = 10 + 2(ab + bc + ca) \)
\( 16 = 10 + 2(ab + bc + ca) \)
Subtract 10 from both sides: \( 6 = 2(ab + bc + ca) \), so \( ab + bc + ca = 3 \).

Step2: Find \( abc \)

We use the identity \( a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) \). Substitute \( a + b + c = 4 \), \( a^2 + b^2 + c^2 = 10 \), \( ab + bc + ca = 3 \), \( a^3 + b^3 + c^3 = 22 \) into it:
\( 22 - 3abc = 4\times(10 - 3) \)
\( 22 - 3abc = 4\times7 \)
\( 22 - 3abc = 28 \)
Subtract 22 from both sides: \( -3abc = 6 \), so \( abc = -2 \).

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

We use the identity \( (a^2 + b^2 + c^2)^2 = a^4 + b^4 + c^4 + 2(a^2b^2 + b^2c^2 + c^2a^2) \). First, we need to find \( a^2b^2 + b^2c^2 + c^2a^2 \). Notice that \( (ab + bc + ca)^2 = a^2b^2 + b^2c^2 + c^2a^2 + 2abc(a + b + c) \). Substitute \( ab + bc + ca = 3 \), \( abc = -2 \), \( a + b + c = 4 \) into it:
\( 3^2 = a^2b^2 + b^2c^2 + c^2a^2 + 2\times(-2)\times4 \)
\( 9 = a^2b^2 + b^2c^2 + c^2a^2 - 16 \)
Add 16 to both sides: \( a^2b^2 + b^2c^2 + c^2a^2 = 25 \).

Now substitute \( a^2 + b^2 + c^2 = 10 \), \( a^2b^2 + b^2c^2 + c^2a^2 = 25 \) into \( (a^2 + b^2 + c^2)^2 = a^4 + b^4 + c^4 + 2(a^2b^2 + b^2c^2 + c^2a^2) \):
\( 10^2 = a^4 + b^4 + c^4 + 2\times25 \)
\( 100 = a^4 + b^4 + c^4 + 50 \)
Subtract 50 from both sides: \( a^4 + b^4 + c^4 = 50 \).

Answer:

\( 50 \)