$x$ is a real number with the property that $x+\tfrac1x = 3$. Let $S_m = x^m + \tfrac{1}{x^m}$. Determine the value of $S_7$.

We can calculate\[x^2 + \dfrac{1}{x^2} = \left(x + \dfrac{1}{x}\right)^2 - 2 = 3^2 -2 = 7.\]Similarly,\[x^3 + \dfrac{1}{x^3} = \left(x + \dfrac{1}{x}\right) \left(x^2 + \dfrac{1}{x^2}\right) - \left(x + \dfrac{1}{x}\right) = 3 \cdot 7 - 3 = 18\]and\[x^4 + \dfrac{1}{x^4} = \left(x^2 + \dfrac{1}{x^2}\right)^2 - 2 = 7^2 - 2 = 47.\]Finally,\[x^7 + \dfrac{1}{x^7} = \left(x^3 + \dfrac{1}{x^3}\right) \left(x^4 + \dfrac{1}{x^4}\right) - \left(x + \dfrac{1}{x}\right) = 18 \cdot 47 - 3 = \boxed{843}.\]
843
Level 5
Intermediate Algebra
Problem: $x$ is a real number with the property that $x+\tfrac1x = 3$. Let $S_m = x^m + \tfrac{1}{x^m}$. Determine the value of $S_7$.

Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the value of $S_7$, we start with the given equation $x + \frac{1}{x} = 3$. We can derive a recurrence relation for $S_m = x^m + \frac{1}{x^m}$ using the identity:

S_m = (x + \frac{1}{x}) S_{m-1} - S_{m-2}.

First, we calculate $S_1$ and $S_2$:

S_1 = x + \frac{1}{x} = 3.

To find $S_2$, we use the square of $S_1$:

S_2 = (x + \frac{1}{x})^2 - 2 = 3^2 - 2 = 9 - 2 = 7.

Now we have:

S_1 = 3,
S_2 = 7.

Next, we apply the recurrence relation to find $S_3$:

S_3 = (x + \frac{1}{x}) S_2 - S_1 = 3 \cdot 7 - 3 = 21 - 3 = 18.

Now we calculate $S_4$:

S_4 = (x + \frac{1}{x}) S_3 - S_2 = 3 \cdot 18 - 7 = 54 - 7 = 47.

Next, we find $S_5$:

S_5 = (x + \frac{1}{x}) S_4 - S_3 = 3 \cdot 47 - 18 = 141 - 18 = 123.

Now we calculate $S_6$:

S_6 = (x + \frac{1}{x}) S_5 - S_4 = 3 \cdot 123 - 47 = 369 - 47 = 322.

Finally, we find $S_7$:

S_7 = (x + \frac{1}{x}) S_6 - S_5 = 3 \cdot 322 - 123 = 966 - 123 = 843.

Thus, the final answer is 843.<|END_OF_LATENT|> The final answer is $843$.