The value is 50.
Explanation (simple)
We are given the sums of the 1st, 2nd, and 3rd powers of three numbers:
- $p_1=a+b+c=4$
- $p_2=a^2+b^2+c^2=10$
- $p_3=a^3+b^3+c^3=22$
Newton’s identities are simple rules that connect these “power sums” $p_k$ to the three elementary symmetric sums:
- $s_1=a+b+c$
- $s_2=ab+bc+ca$
- $s_3=abc$
For three numbers, knowing $p_1,p_2,p_3$ lets us find $s_2$ and $s_3$, and then compute $p_4=a^4+b^4+c^4$.
Steps (calculation)
- Use the identity for $p_2$:
$$p_2=s_1^2-2s_2$$
Substitute $p_2=10$ and $s_1=4$:
$$10=4^2-2s_2\implies10=16-2s_2\implies2s_2=6\implies s_2=3.$$
- Use the identity for $p_3$:
$$p_3=s_1p_2-s_2p_1+3s_3$$
Substitute $p_3=22$, $s_1=4$, $p_2=10$, $s_2=3$, $p_1=4$:
$$22=4\cdot10-3\cdot4+3s_3\implies22=40-12+3s_3\implies3s_3=-6\implies s_3=-2.$$
- Now compute $p_4$ using the three-variable identity:
$$p_4=s_1p_3-s_2p_2+s_3p_1$$
Substitute values:
$$p_4=4\cdot22-3\cdot10+(-2)\cdot4=88-30-8=50.$$
Therefore $a^4+b^4+c^4=50$.
Quick intuition: Newton’s identities let you trade between sums of powers and the basic symmetric combinations ($s_1,s_2,s_3$). For three numbers, the first three power sums determine those symmetric sums, which then give any higher power sum you need.