Using Symmetry: Practice

Recover the product from $x^2 + y^2 = s^2 - 2p$.

$20 = 6^2 - 2p \implies 2p = 16 \implies p = 8$

So $xy = 8$.

Use $x^3 + \frac{1}{x^3} = k^3 - 3k$ with $k = 4$.

$4^3 - 3(4) = 64 - 12 = 52$

The correction term $3k$ subtracts $12$.

Since $(x - y)^2 = s^2 - 4p$ and a real square is never negative:

$s^2 - 4p \ge 0 \implies s^2 \ge 4p$

Equality gives a repeated value $x = y$.

Square $k$ and subtract the middle term, which is $2$ because $x \cdot \frac{1}{x} = 1$.

$\left(x + \tfrac{1}{x}\right)^2 = x^2 + \tfrac{1}{x^2} + 2 \implies x^2 + \tfrac{1}{x^2} = k^2 - 2$

Use $(x - y)^2 = s^2 - 4p$ with $s = 7$ and $p = 10$.

$7^2 - 4(10) = 49 - 40 = 9$

So $x$ and $y$ differ by $\sqrt{9} = 3$.

The axis is $x = -\frac{b}{2a}$ with $a = 1$ and $b = -6$.

$x = -\frac{-6}{2(1)} = 3$

The two roots $1$ and $5$ are mirror images about $x = 3$.

The two numbers add to the sum, so subtract the known one.

$9 - 4 = 5$

The other number is $5$.

Use $x^3 + y^3 = s^3 - 3ps$ with $s = 4$ and $p = 1$.

$4^3 - 3(1)(4) = 64 - 12 = 52$

The correction term carries a factor of $s = 4$.

The roots of a palindromic equation come in reciprocal pairs.

$3 \to \tfrac{1}{3}, \qquad 5 \to \tfrac{1}{5}$

So $\frac{1}{3}$ and $\frac{1}{5}$ must also be roots.

Build the quadratic $t^2 - st + p$ with $s = -5$ and $p = 6$.

$t^2 + 5t + 6 = (t + 2)(t + 3) = 0$

The roots are $-2$ and $-3$, so $\{x, y\} = \{-2, -3\}$.

Squaring erases the sign that the swap flips.

$(x - y)^2 = (y - x)^2 = s^2 - 4p$

So $(x - y)^2$ is symmetric even though $x - y$ is not.

Use $x^2 + y^2 = s^2 - 2p$ with $s = 8$ and $p = 15$.

$8^2 - 2(15) = 64 - 30 = 34$