2020 AMC 10A Problem 14

The numbers $x$ and $y$ themselves are messy (they involve square roots), but we know their sum and product. So rewrite everything in terms of $x + y$ and $xy$.

1. Combine the fractions. The common bottom is $x^2 y^2$:
$$\frac{x^3}{y^2} + \frac{y^3}{x^2} = \frac{x^3 \cdot x^2 + y^3 \cdot y^2}{x^2 y^2} = \frac{x^5 + y^5}{x^2 y^2}.$$
2. The easy parts. $x + y = 4$, and $x^2 y^2 = (xy)^2 = (-2)^2 = 4$. So the whole expression is
$$4 + \frac{x^5 + y^5}{4}.$$
3. Find $x^2 + y^2$. Square the sum: $(x + y)^2 = x^2 + 2xy + y^2$, so
$$x^2 + y^2 = 16 - 2(-2) = 20.$$
4. Find $x^3 + y^3$. Multiply: $(x + y)(x^2 + y^2) = x^3 + y^3 + xy(x + y)$. So
$$x^3 + y^3 = 4 \times 20 - (-2)(4) = 80 + 8 = 88.$$
5. Find $x^5 + y^5$. Multiply the last two results in the same way:
$$(x^2 + y^2)(x^3 + y^3) = x^5 + y^5 + x^2 y^2 (x + y).$$
So $x^5 + y^5 = 20 \times 88 - 4 \times 4 = 1760 - 16 = 1744$.
6. Finish. $4 + \frac{1744}{4} = 4 + 436 = 440$.

The answer is $\boxed{\textbf{(D)}\ 440}$.

Any expression that stays the same when you swap $x$ and $y$ can be written using only $x + y$ and $xy$. Multiplying two such sums, like $(x^2 + y^2)(x^3 + y^3)$, gives the bigger power you want plus a leftover "cross" part that is easy to subtract. So you never need the actual values of $x$ and $y$.

Build the power sums one at a time. Because $x^2 = 4x + 2$ (from $x^2 - 4x - 2 = 0$, which $x$ and $y$ both satisfy), each power sum is $4$ times the previous one plus $2$ times the one before that:
$$x^3 + y^3 = 4(20) + 2(4) = 88, \quad x^4 + y^4 = 4(88) + 2(20) = 392, \quad x^5 + y^5 = 4(392) + 2(88) = 1744.$$
Then the answer is $4 + 1744 \div 4 = 440$.