Using Symmetry: Practice

Swap $x$ and $y$ in each expression and see which one is altered.

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

The sum, the product, and $x^2 + y^2$ are unchanged, but $x - y$ turns into its own negative, so it is not symmetric.

A symmetric expression is unchanged when $x$ and $y$ trade places.

$x^2 + y^2 \ \longrightarrow \ y^2 + x^2$

Only $x^2 + y^2$ reads the same after the swap; $x - y$, $x^2 y$, and $x y^2$ all change.

Use $x^2 + y^2 = s^2 - 2p$ with $s = 4$ and $p = 3$.

$4^2 - 2(3) = 16 - 6 = 10$

Adding the cross term instead of subtracting would give the wrong value $22$.

Add the two fractions over a common denominator.

$\frac{1}{x} + \frac{1}{y} = \frac{x + y}{xy} = \frac{s}{p} = \frac{5}{4}$

The sum goes on top and the product on the bottom.

Sum $5$ and product $6$ give the quadratic $t^2 - 5t + 6 = 0$.

$(t - 2)(t - 3) = 0$

The roots are $2$ and $3$, so the pair is $\{2, 3\}$.

Square $k = x + \frac{1}{x}$ and subtract the middle term $2$.

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

With $k = 3$ this is $9 - 2 = 7$.

A palindromic equation has coefficients that mirror end to end.

$1, \ 2, \ 3, \ 2, \ 1$

Only $x^4 + 2x^3 + 3x^2 + 2x + 1 = 0$ has this mirror-image list of coefficients.

In a palindromic equation the roots come in reciprocal pairs, so if $r$ is a root then $\frac{1}{r}$ is a root.

$r = 4 \implies \frac{1}{r} = \frac{1}{4}$

So $\frac{1}{4}$ must also be a root.

Use $(x - y)^2 = (x + y)^2 - 4xy$.

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

Even though $x - y$ is not symmetric, its square is, and it reduces to $s$ and $p$.

Use $x^3 + y^3 = s^3 - 3ps$ with $s = 3$ and $p = 2$.

$3^3 - 3(2)(3) = 27 - 18 = 9$

The correction term $3ps$ carries a factor of $s$, so it is $18$, not $6$.

The product $p$ is read straight from $xy$.

$p = xy = 15$

The number $8$ is the sum $s$, not the product.

The reduction of the sum of squares is $x^2 + y^2 = s^2 - 2p$.

$s^2 - 2p = x^2 + y^2$

So the quantity equal to $s^2 - 2p$ is $x^2 + y^2$.