Q.Simplify by rationalising the denominator: $\dfrac{7+\sqrt6}{3-\sqrt2}$.

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. …