The correct answer is: $50$
Explanation (simple)
We use simple algebra identities that relate sums of powers to the basic symmetric sums:
- Let $s_1=a+b+c$, $s_2=ab+bc+ca$, $s_3=abc$.
- We are given $p_1=a+b+c=4$, $p_2=a^2+b^2+c^2=10$, $p_3=a^3+b^3+c^3=22$.
- From these we find $s_2$ and $s_3$, then use a formula for the fourth powers.
Steps
- Use the identity $a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)$ to get $s_2$:
$$
10 = 4^2 – 2s_2 \quad\Rightarrow\quad 10=16-2s_2 \quad\Rightarrow\quad s_2=3.
$$
- Use the cube identity $a^3+b^3+c^3=(a+b+c)^3-3(a+b+c)(ab+bc+ca)+3abc$ to get $s_3$:
$$
22 = 4^3 – 3\cdot4\cdot3 + 3s_3 \quad\Rightarrow\quad 22=64-36+3s_3 \quad\Rightarrow\quad 3s_3=-6 \quad\Rightarrow\quad s_3=-2.
$$
- For three variables, Newton’s identity gives the fourth power sum:
$$
a^4+b^4+c^4 = s_1 p_3 – s_2 p_2 + s_3 p_1.
$$
Plug in the numbers:
$$
p_4 = 4\cdot22 – 3\cdot10 + (-2)\cdot4 = 88 – 30 – 8 = 50.
$$
So $a^4+b^4+c^4 = 50$.