n-th roots
Definition
The n-th root of $a$ is $b$ if $b^{n}=a$.
$$\sqrt[n]{a}=b\iff b^{n}=a$$
$n$ is the index, $a$ the radicand and $b$ the result.
Examples:
$$\sqrt[3]{27}=3\iff 3^{3}=27 \qquad \sqrt[3]{8}=2\iff 2^{3}=8$$
Root – power relation
With $m\in\mathbb{Z}$ and $n\in\mathbb{N}$, $n>1$:
$$\sqrt[n]{a^{m}}=a^{m/n}$$
Examples:
$$\sqrt[5]{2^{4}}=2^{4/5}\qquad \sqrt{2}=2^{1/2}\qquad \sqrt[4]{2^{6}}=2^{6/4}=2^{3/2}=\sqrt{2^{3}}$$
Simplifying and amplifying (index – exponent)
Multiplying (or dividing) the index and the exponent by the same number does not change the root:
$$\sqrt[n\cdot m]{a^{p\cdot m}}=\sqrt[n]{a^{p}}$$
Example:
$$\sqrt[4]{1024}=\sqrt[4]{2^{10}}=\sqrt{2^{5}}$$
(We divide the index $4$ and the exponent $10$ by $2$.)
Taking factors out and putting factors in
$$\sqrt[n]{a^{p\cdot n}\cdot b}=a^{p}\cdot\sqrt[n]{b}$$
Examples:
$$\sqrt[3]{3^{5}}=\sqrt[3]{3^{3}\cdot 3^{2}}=3\cdot\sqrt[3]{3^{2}}$$
$$5\cdot\sqrt[4]{5^{3}}=\sqrt[4]{5^{4}\cdot 5^{3}}=\sqrt[4]{5^{7}}$$
To take factors out, factor the radicand into primes. Example from the board:
$$12960=2^{5}\cdot 3^{4}\cdot 5$$
$$\sqrt[4]{12960}=\sqrt[4]{2^{5}\cdot 3^{4}\cdot 5}=2\cdot 3\cdot\sqrt[4]{2\cdot 5}=6\cdot\sqrt[4]{10}$$
$$\sqrt[3]{32}=\sqrt[3]{2^{5}}=2\cdot\sqrt[3]{2^{2}}$$