If $\left( r + \frac{1}{r} \right)^2 = 3,$ then find $r^3 + \frac{1}{r^3}.$
Expanding, we get $r^2 + 2 + \frac{1}{r^2} = 3,$ so
\[r^2 - 1 + \frac{1}{r^2} = 0.\]Then
\[r^3 + \frac{1}{r^3} = \left( r + \frac{1}{r} \right) \left( r^2 - 1 + \frac{1}{r^2} \right) = \boxed{0}.\]
0
Level 3
Intermediate Algebra
Problem: If $\left( r + \frac{1}{r} \right)^2 = 3,$ then find $r^3 + \frac{1}{r^3}.$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve for \( r^3 + \frac{1}{r^3} \), we start with the given equation \( \left( r + \frac{1}{r} \right)^2 = 3 \). 

First, we expand the left side:
\[
\left( r + \frac{1}{r} \right)^2 = r^2 + 2 + \frac{1}{r^2}.
\]
Setting this equal to 3 gives:
\[
r^2 + 2 + \frac{1}{r^2} = 3.
\]
Subtracting 2 from both sides results in:
\[
r^2 + \frac{1}{r^2} = 1.
\]

Next, we use the identity for \( r^3 + \frac{1}{r^3} \):
\[
r^3 + \frac{1}{r^3} = \left( r + \frac{1}{r} \right) \left( r^2 - 1 + \frac{1}{r^2} \right).
\]
We already know \( r + \frac{1}{r} \) can be found from \( \left( r + \frac{1}{r} \right)^2 = 3 \), which implies:
\[
r + \frac{1}{r} = \sqrt{3} \text{ or } -\sqrt{3}.
\]

Now, we need \( r^2 - 1 + \frac{1}{r^2} \). We can express this as:
\[
r^2 + \frac{1}{r^2} - 1 = 1 - 1 = 0.
\]

Thus, substituting back into the identity gives:
\[
r^3 + \frac{1}{r^3} = \left( r + \frac{1}{r} \right) \cdot 0 = 0.
\]

Therefore, the final answer is \( 0 \).<|END_OF_LATENT|> The final answer is $0$.