Q.Simplify:

Integer exponents. $a^n=a\cdot a\cdots a$ ($n$ times); $a^{-m}=1/a^m$; $a^ma^n=a^{m+n}$; $a^m/a^n=a^{m-n}$.

$n$th roots (radicals). For $n$ even, $b>0$ has a unique positive $n$th root $b^{1/n}$ (no real root exists for $b<0$); for $n$ odd, every real $b$ has a unique real $n$th root. $n=2$ gives the square root, $n=3$ the cube root. Crucially, $\sqrt{a^2}=|a|$, never plain $a$ -- more generally $(a^n)^{1/n}=|a|$ if $n$ even, $=a$ if $n$ odd.

Rational exponents. For $a>0$ and $r=m/n$ ($\gcd(m,n)=1$): $a^{m/n}=(a^{1/n})^m$; all the integer-exponent laws extend to rational exponents wherever every term involved is actually defined (e.g. $(-49)^{3/2}$ is undefined in the reals, since $(-49)^{1/2}$ isn't real).

Simplifying compound radical/exponent expressions typically means rewriting every base as a perfect power of a common small integer (e.g. $125=5^3$, $256=2^8$, $27=3^3$) before applying the exponent laws, rather than computing large numbers directly.

Rationalising with a conjugate. For rational $u,v,b$ with $b$ a non-square rational, $(u+v\sqrt b)(u-v\sqrt b)=u^2-bv^2$ is rational -- so multiplying a fraction's numerator and denominator by the denominator's conjugate clears the surd from the bottom. The same idea (with $u\sqrt a\pm v\sqrt b$) extends to two different surds. Sums of several such rationalised unit fractions frequently telescope -- most of the surd terms cancel between consecutive terms, leaving a simple rational total.

Un-nesting a double radical. $\sqrt{p-q\sqrt d}$ can sometimes be written as $a-b\sqrt d$ (rational $a,b$) by squaring, matching the rational part ($a^2+b^2d=p$) and the surd part ($2ab=q$) separately, and solving the resulting system -- though not every nested radical un-nests into rational $a,b$.