Using Symmetry: Practice

Use $x^3 + y^3 = s^3 - 3ps$ with $s = 5$ and $p = 2$.

$5^3 - 3(2)(5) = 125 - 30 = 95$

The correction term $3ps = 30$ carries the factor of $s$.

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

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

Dividing by $x^2$ loses nothing only if $x = 0$ is not a root, which holds when the constant term is nonzero.

$2x^4 - 9x^3 + 14x^2 - 9x + 2 = 2 \ne 0 \ \text{at} \ x = 0$

So the division is safe here.

Use $x^3 + \frac{1}{x^3} = k^3 - 3k$ with $k = 5$.

$5^3 - 3(5) = 125 - 15 = 110$

Use $(x - y)^2 = s^2 - 4p = 49 - 40 = 9$, then take the square root.

$|x - y| = \sqrt{9} = 3$

Setting $k = x + \frac{1}{x}$ with product $1$, a real $x$ needs $k^2 \ge 4(1)$.

$k^2 \ge 4 \implies |k| \ge 2$

Only $2.5$ satisfies $|k| \ge 2$, so it is the only reachable value listed.

The two roots are conjugates, so their sum and product are rational: sum $= 4$, product $= (2)^2 - (\sqrt{3})^2 = 1$.

$t^2 - 4t + 1 = 0$

Both totals come from $s$ and $p$ without multiplying the roots out separately.

Grouping the reciprocal terms and substituting $u$ halves the degree.

$\text{degree } 4 \ \longrightarrow \ \text{degree } 2$

A quartic becomes a quadratic in $u$, which is why it can be solved by factoring.

Build $t^2 - st + p$ with $s = 1$ and $p = -6$.

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

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

Use $(x - y)^2 = s^2 - 4p = 100 - 84 = 16$.

$|x - y| = \sqrt{16} = 4$

The two numbers are $3$ and $7$, which indeed differ by $4$.

Use $x^2 + \frac{1}{x^2} = k^2 - 2$ and solve for $k$.

$k^2 - 2 = 23 \implies k^2 = 25 \implies k = 5$

Taking the positive value gives $k = 5$.

Check $s^2$ against $4p$: $16 < 20$, so $(x - y)^2 = s^2 - 4p = -4 < 0$.

$s^2 = 16 < 20 = 4p$

No real pair exists; the two numbers form a complex conjugate pair.