2021 AMC 10B Problem 15

Step 1: notice the pattern in the question.

1. The exponents $11$, $7$, $3$ go down by $4$ each time, and the middle term has a $7$. That hints that we want to know about $x^4$, and in particular that something equals $7$.

Step 2: square to climb to $x^4$.

2. Square $x + \frac1x = \sqrt5$. The middle term is $2 \cdot x \cdot \frac1x = 2$:
$$x^2 + 2 + \frac{1}{x^2} = 5, \qquad \text{so} \qquad x^2 + \frac{1}{x^2} = 3.$$
3. Square again, the same way:
$$x^4 + 2 + \frac{1}{x^4} = 9, \qquad \text{so} \qquad x^4 + \frac{1}{x^4} = 7.$$
4. There is the $7$.

Step 3: turn it into the expression we want.

5. Multiply both sides of $x^4 + \frac{1}{x^4} = 7$ by $x^7$:
$$x^{11} + x^3 = 7x^7.$$
6. Move $7x^7$ to the left:
$$x^{11} - 7x^7 + x^3 = 0.$$

The answer is $\boxed{\textbf{(B)}\ 0}$.

From $x + \frac1x$ you can square your way to $x^2 + \frac1{x^2}$, then $x^4 + \frac1{x^4}$, and so on, without ever finding $x$. When the exponents in the target are evenly spaced, the middle one is where to multiply, so that both ends of the target appear together. Look at the numbers in the question for hints: the $7$ here is exactly $x^4 + \frac1{x^4}$.

Square the given relation:
$$x^2 + 2 + \frac{1}{x^2} = 5 \quad\Longrightarrow\quad x^2 + \frac{1}{x^2} = 3.$$
Square again:
$$x^4 + 2 + \frac{1}{x^4} = 9 \quad\Longrightarrow\quad x^4 + \frac{1}{x^4} = 7.$$

The coefficient $7$ in the problem is the signal. Multiply the last equation by $x^4$:
$$x^8 + 1 = 7x^4 \quad\Longrightarrow\quad x^8 - 7x^4 + 1 = 0.$$
Multiplying by $x^3$ gives exactly the target:
$$x^{11} - 7x^7 + x^3 = x^3\left(x^8 - 7x^4 + 1\right) = 0.$$

The answer is $\boxed{\textbf{(B)}\ 0}$.