2021 AMC 10B Problem 15

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 $\boxed{\textbf{(B)}\ 0}$.

Expressions in $x + \frac{1}{x}$ generate $x^{2^k} + \frac{1}{x^{2^k}}$ by repeated squaring, and each such relation, once cleared of fractions, is a polynomial that $x$ satisfies. The target $x^{11} - 7x^7 + x^3$ has exponents in arithmetic progression with common difference $4$, so it is $x^3$ times a polynomial in $x^4$; match that polynomial against $x^4 + \frac{1}{x^4}$.

From $x^2 - \sqrt{5}\,x + 1 = 0$ one can reduce powers step by step, but that involves $\sqrt{5}$ at every stage. Noticing that all three exponents differ by $4$ and factoring out $x^3$ first is the faster route.

• Solving for x explicitly (x = (√5 ± 1)/2) and trying to raise it to the 11th power, which is slow and error-prone.
• Forgetting the middle term when squaring, i.e. writing $x^2 + \frac{1}{x^2} = 5$ instead of $3$.