Using Symmetry: Free Response

Read off $s = 8$ and $p = 3$ from the given system, then apply the reduction for the sum of squares.

$x^2 + y^2 = s^2 - 2p = 8^2 - 2(3) = 64 - 6 = 58.$

Neither $x$ nor $y$ had to be found individually.

Apply the reduction for the sum of cubes with $s = 8$ and $p = 3$, keeping the factor of $s$ in the correction term.

$x^3 + y^3 = s^3 - 3ps = 8^3 - 3(3)(8) = 512 - 72 = 440.$

Expand the cube of the sum in full.

$(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3 = x^3 + y^3 + 3xy(x+y).$

The leftover piece is $3xy(x+y)$, not merely $3xy$: the factor $(x+y)$ is still attached. Since $xy = p$ and $x+y = s$, that piece is $3ps$, and rearranging gives $x^3+y^3 = s^3 - 3ps$. Dropping the trailing $(x+y)$ is exactly how the shorter, incorrect form arises.

The two unknowns are the roots of the monic quadratic built from their sum and product.

$t^2 - st + p = 0 \quad\Longrightarrow\quad t^2 - 9t + 14 = 0.$

Factor: $(t-2)(t-7) = 0$, so the roots are $t = 2$ and $t = 7$. The pair of numbers with sum $9$ and product $14$ is $\{2, 7\}$.

Swap $x$ and $y$ in the original system: $x+y=9$ becomes $y+x=9$, the same equation, and $xy=14$ becomes $yx=14$, also the same equation.

$x+y=9,\ xy=14 \quad\longleftrightarrow\quad y+x=9,\ yx=14.$

Since the system genuinely cannot tell $x$ apart from $y$, any solution $(x,y)$ it has must sit alongside its swapped twin $(y,x)$ as an equally valid solution. Both orderings from part A satisfy the system, so the honest report is the unordered pair, not a claim about which number is $x$.

Check $x=0$ first: substituting gives the constant term, $3 \neq 0$, so $x=0$ is not a root and dividing by $x^2$ is safe. Divide every term by $x^2$:

$3x^2 - 4x - 14 - \frac{4}{x} + \frac{3}{x^2} = 0.$

Group the mirrored terms and substitute $u = x + \frac{1}{x}$, using $x^2 + \frac{1}{x^2} = u^2 - 2$:

$3\left(x^2+\frac{1}{x^2}\right) - 4\left(x+\frac{1}{x}\right) - 14 = 3(u^2-2) - 4u - 14 = 3u^2 - 4u - 20 = 0.$

Apply the quadratic formula to $3u^2-4u-20=0$.

$u = \frac{4 \pm \sqrt{16 + 240}}{6} = \frac{4 \pm 16}{6}.$

So $u = \frac{20}{6} = \frac{10}{3}$ or $u = \frac{-12}{6} = -2$.

For $u = \frac{10}{3}$, clear the fraction:

$x + \frac{1}{x} = \frac{10}{3} \quad\Longrightarrow\quad 3x^2 - 10x + 3 = 0 \quad\Longrightarrow\quad (3x-1)(x-3)=0,$

giving $x = 3$ or $x = \frac{1}{3}$. For $u = -2$:

$x + \frac{1}{x} = -2 \quad\Longrightarrow\quad x^2 + 2x + 1 = 0 \quad\Longrightarrow\quad (x+1)^2 = 0,$

giving $x = -1$ as a double root. The four roots (with multiplicity) are $3$, $\frac{1}{3}$, $-1$, and $-1$.

In a palindromic equation, if $r$ is a root then so is $\frac{1}{r}$. Check the pair found first:

$3 \times \frac{1}{3} = 1,$

confirming $3$ and $\frac{1}{3}$ are genuine reciprocals of each other. Now check the repeated root:

$\frac{1}{-1} = -1.$

Since $-1$ is its own reciprocal, the reciprocal pairing does not require a second, different root to accompany it; the pairing simply closes up on itself. The double root is that self-paired value appearing with multiplicity two, which is exactly what the pairing predicts rather than a violation of it.

Compute the expression as given, then swap the letters and compute again.

$x - y = 9 - 4 = 5, \qquad y - x = 4 - 9 = -5.$

The two values disagree, so for this one pair the swap changes the expression's value: the claim does not survive this test.

Swap every $x$ for $y$ and every $y$ for $x$ in the expression itself.

$x - y \ \longrightarrow\ y - x = -(x-y).$

The swapped expression equals the original, $-(x-y) = x-y$, only when $2(x-y)=0$, that is, only when $x=y$. For any pair with $x \neq y$, such as the one tested above, the swap changes the value, so $x-y$ fails the symmetry test in general, not merely for one unlucky choice.

Square the swapped expression from part B.

$(y-x)^2 = \big(-(x-y)\big)^2 = (x-y)^2.$

Whatever sign the swap attaches to $x-y$, squaring removes it, since a number and its negative have the same square. That is exactly why $(x-y)^2$, unlike $x-y$ itself, is symmetric, and it reduces to the two symmetric building blocks the same way every other symmetric expression in this lesson does, $(x-y)^2 = s^2-4p$.

The quadratic formula gives both roots at once, differing only in the sign in front of the square root.

$r_1 = \frac{-b-\sqrt{b^2-4ac}}{2a}, \qquad r_2 = \frac{-b+\sqrt{b^2-4ac}}{2a}.$

Add the two root expressions first.

$r_1 + r_2 = \frac{-b-\sqrt{b^2-4ac}}{2a} + \frac{-b+\sqrt{b^2-4ac}}{2a} = \frac{-2b}{2a}.$

The square-root terms are the same quantity with opposite signs, so they cancel exactly. Dividing by $2$ for the midpoint:

$\frac{r_1+r_2}{2} = \frac{1}{2}\cdot\frac{-2b}{2a} = -\frac{b}{2a}.$

Rewrite each root as the midpoint plus or minus the same offset.

$r_1 = -\frac{b}{2a} - \frac{\sqrt{b^2-4ac}}{2a}, \qquad r_2 = -\frac{b}{2a} + \frac{\sqrt{b^2-4ac}}{2a}.$

Both roots sit the same distance, $\frac{\sqrt{b^2-4ac}}{2a}$, from $-\frac{b}{2a}$, one below and one above it, which is exactly what it means to be mirror images across that vertical line. Nowhere in this argument did a specific value of $a$, $b$, or $c$ get used, only the general quadratic formula, so the same conclusion holds for every quadratic with two real roots, not merely the one worked here.