Q.Solve: $8x^{\frac{3}{2n}}-8x^{\frac{-3}{2n}}=63$

When an equation's coefficients hide a spottable pattern — even powers only, coefficients summing to zero, matching odd/even sums, a partly-factored shape, or a disguised non-polynomial form — a substitution collapses it to a lower-degree (usually quadratic) equation.

Only even powers present. A degree-$2n$ equation with every odd-power coefficient $=0$ becomes a genuine degree-$n$ equation under $y=x^2$; each root $y_r$ then gives up to two $x$-roots via $x=\pm\sqrt{y_r}$. (E.g. $x^4-9x^2+20=0 \to y^2-9y+20=0=(y-4)(y-5)$, giving $x=\pm2,\pm\sqrt5$.)

Coefficients sum to zero. The coefficient sum is exactly $P(1)$, so a zero sum means $1$ is always a root — an immediate first factor to divide out.

Odd-power sum equals even-power sum. This is exactly the "coefficients of $P(-x)$ sum to zero" condition in disguise, so $-1$ is always a root.

Partly-factored quartics $(ax+b)(cx+d)(px+q)(rx+s)+k=0$ can often be re-paired so two pairs of factors expand to quadratics sharing the same $x^2$- and $x$-coefficient; substituting $y=$ that shared quadratic expression collapses the quartic to a quadratic in $y$.

Genuinely non-polynomial equations (radical equations, or trigonometric equations that are secretly polynomial in $\sin x$ or $\cos x$) become real polynomial equations after the right substitution — but three honest cautions apply: not every derived root solves the original equation (check back, since squaring especially can manufacture extraneous roots); the original equation can have infinitely many solutions (e.g. every $\cos x=\tfrac12$ solution, $x=2n\pi\pm\tfrac\pi3$); or it can have none at all if the derived polynomial's roots fall outside the valid range (e.g. $\cos x=4$ is impossible).

Worked illustration (zero coefficient sum). $x^3-3x^2-33x+35=0$: coefficients sum to $0$, so $1$ is a root; dividing by $(x-1)$ leaves $x^2-2x-35=(x-7)(x+5)$. Roots: $1,7,-5$.