Let $ a$, $ b$, $ c$ be nonzero real numbers such that $ a+b+c=0$ and $ a^3+b^3+c^3=a^5+b^5+c^5$. Find the value of
$ a^2+b^2+c^2$.
From the factorization
\[a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc),\]we know that $a^3 + b^3 + c^3 = 3abc.$

Since $a + b + c = 0,$ $c = -a - b,$ so
\begin{align*}
a^5 + b^5 + c^5 &= a^5 + b^5 - (a + b)^5 \\
&= -5a^4 b - 10a^3 b^2 - 10a^2 b^3 - 5ab^4 \\
&= -5ab(a^3 + 2a^2 b + 2ab^2 + b^3) \\
&= -5ab[(a^3 + b^3) + (2a^2 b + 2ab^2)] \\
&= -5ab[(a + b)(a^2 - ab + b^2) + 2ab(a + b)] \\
&= -5ab(a + b)(a^2 + ab + b^2) \\
&= 5abc(a^2 + ab + b^2),
\end{align*}so
\[3abc = 5abc(a^2 + ab + b^2).\]Since $a,$ $b,$ $c$ are all nonzero, we can write
\[a^2 + ab + b^2 = \frac{3}{5}.\]Hence,
\begin{align*}
a^2 + b^2 + c^2 &= a^2 + b^2 + (a + b)^2 \\
&= a^2 + b^2 + a^2 + 2ab + b^2 \\
&= 2a^2 + 2ab + 2b^2 \\
&= 2(a^2 + ab + b^2) = \boxed{\frac{6}{5}}.
\end{align*}
\frac{6}{5}
Level 5
Intermediate Algebra
Problem: Let $ a$, $ b$, $ c$ be nonzero real numbers such that $ a+b+c=0$ and $ a^3+b^3+c^3=a^5+b^5+c^5$. Find the value of
$ a^2+b^2+c^2$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the problem, we start with the given conditions: \( a + b + c = 0 \) and \( a^3 + b^3 + c^3 = a^5 + b^5 + c^5 \). 

Using the identity for the sum of cubes, we can express \( a^3 + b^3 + c^3 \) in terms of \( a + b + c \) and \( ab + ac + bc \):
\[
a^3 + b^3 + c^3 = 3abc \quad \text{(since } a + b + c = 0\text{)}.
\]

Next, we need to express \( a^5 + b^5 + c^5 \). We can use the recurrence relation for power sums:
\[
P_n = a^n + b^n + c^n,
\]
where \( P_1 = a + b + c = 0 \), \( P_2 = a^2 + b^2 + c^2 \), and \( P_3 = 3abc \). The recurrence relation is given by:
\[
P_n = (a + b + c)P_{n-1} - (ab + ac + bc)P_{n-2} + abcP_{n-3}.
\]
Since \( a + b + c = 0 \), this simplifies to:
\[
P_n = -(ab + ac + bc)P_{n-2} + abcP_{n-3}.
\]

Using this, we can compute \( P_4 \) and \( P_5 \):
\[
P_4 = -(ab + ac + bc)P_2 + abcP_1 = -(ab + ac + bc)P_2,
\]
\[
P_5 = -(ab + ac + bc)P_3 + abcP_2 = -3abc(ab + ac + bc) + abcP_2.
\]

Setting \( P_3 = P_5 \) gives us:
\[
3abc = -3abc(ab + ac + bc) + abcP_2.
\]
Dividing through by \( abc \) (since \( a, b, c \) are nonzero):
\[
3 = -3(ab + ac + bc) + P_2.
\]
Rearranging yields:
\[
P_2 = 3 + 3(ab + ac + bc).
\]

Next, we can find \( ab + ac + bc \) using the identity:
\[
(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + ac + bc).
\]
Since \( a + b + c = 0 \):
\[
0 = a^2 + b^2 + c^2 + 2(ab + ac + bc).
\]
Thus,
\[
ab + ac + bc = -\frac{1}{2}(a^2 + b^2 + c^2).
\]

Substituting this into the equation for \( P_2 \):
\[
P_2 = 3 + 3\left(-\frac{1}{2}(a^2 + b^2 + c^2)\right) = 3 - \frac{3}{2}P_2.
\]
Let \( x = P_2 \):
\[
x + \frac{3}{2}x = 3 \implies \frac{5}{2}x = 3 \implies x = \frac{6}{5}.
\]

Thus, the value of \( a^2 + b^2 + c^2 \) is \( \frac{6}{5} \).<|END_OF_LATENT|> The final answer is $\frac{6}{5}$.