Radical Cancellation and Fractional Power Representation
Cancellation Rules
Suppose \(n\) is an even natural number. Then
$$\sqrt[n]{x^n}=(\sqrt[n]{x})^n=|x|$$
Moral of the Story
Whether you are taking a root of a power or a power of the root, if the “root power” and the exponent is the same, then you are left with the absolute value of what was inside both.
\(\sqrt{x^2}=|x|\), i.e. the square root cancels the square, leaving you with absolute value of the inside, i.e. \(x\).
\(\sqrt[4]{x^4}=|x|\), i.e. the 4th root cancels the 4th power, leaving you with the absolute value of the inside expression, i.e. \(x\).
\((\sqrt{x-1})^2=|x-1|\), i.e. the square on the outside cancels out the square root on the inside, leaving you with the absolute value of the inside expression, namely \(|x-1|\).
\((\sqrt[6]{x^2-1})^6=|x^2-1|\), i.e. the 6th power on the outside kills the 6th root on the inside, leaving you with the absolute value of the inside expression.
Take a look at the expression \(\sqrt{x^2}\).
If we were to replace \(x\) with a positive number, such as \(5\), notice
\(\sqrt{(5)^2}=\sqrt{25}=5\).
Similarly, if you plugged in, say, \(x=3\), you get
\(\sqrt{(3)^2}=\sqrt{9}=3\).
The same exact thing would happen if you plugged in any positive number \(x=p\):
\(\sqrt{(p)^2}=\sqrt{p^2}=p\).
Now what happens if you instead plug in negative numbers?
Lets replace \(x\) with \(-2\):
\(\sqrt{(-2)^2}=\sqrt{4}=2\).
Similarly, if we have \(x=-5\) instead, notice
\(\sqrt{(-5)^2}=\sqrt{25}=5\).
The same thing will happen if you plug in any negative number \(-p\):
\(\sqrt{(-p)^2}=sqrt{p^2}=p\).
In any of these cases , it seems that the square is cancelling the negative out in front of the number (because negative times a negative is positive, and squaring a number is the same as multiplying that number by itself). Then, when the square root is taken, you are get the positive version of the number you plugged in!
Thus, we have
\(\sqrt{x^2}=|x|\) because the power \(2\) makes negative numbers positive, and the square root undoes the “squaring”, which leaves you with the nonnegative version of whatever is plugged in for \(x\).
This idea generalizes to other even powers and roots, such as \(4,6,8,10,\) etc. as well.
Suppose \(n\) is an odd natural number. Then
$$\sqrt[n]{x^n}=(\sqrt[n]{x})^n=x$$
Moral of the Story
Whether you are taking a root of a power or a power of the root, if the “root power” and the exponent is the same, then you are left with what was inside both.
\(\sqrt{x^3}=x\), i.e. the square root cancels the square, leaving you with absolute value of the inside, i.e. \(x\).
\(\sqrt[5]{3^5}=3\), i.e. the 5th root cancels the 5th power, leaving you with what you started with: \(3\).
\((\sqrt[3]{x-1})^3=x-1\), i.e. the third power on the outside cancels out the cube root on the inside, leaving you with the absolute value of the inside expression, namely \(x-1\).
\((\sqrt[13]{x^2-1})^{13}=x^2-1\), i.e. the 13th power on the outside kills the 13th root on the inside, leaving you with the absolute value of the inside expression.
Take a look at the expression \(\sqrt{x^3}\).
If we were to replace \(x\) with a positive number, such as \(5\), notice
\(\sqrt[3]{(5)^3}=\sqrt[3]{125}=5\).
Similarly, if you plugged in, say, \(x=3\), you get
\(\sqrt[3]{(3)^3}=\sqrt[3]{27}=3\).
The same exact thing would happen if you plugged in any positive number \(x=p\):
\(\sqrt[3]{(p)^3}=\sqrt{p^3}=p\).
Now what happens if you instead plug in negative numbers?
Lets replace \(x\) with \(-2\):
\(\sqrt[3]{(-2)^3}=\sqrt{-8}=-2\). (Because raising a negative number to an odd power gives you a negative number; same with taking odd roots of negatives.)
Similarly, if we have \(x=-5\) instead, notice
\(\sqrt[3]{(-5)^3}=\sqrt{-125}=-5\).
The same thing will happen if you plug in any negative number \(-p\):
\(\sqrt[3]{(-p)^3}=\sqrt[3]{-p^3}=-p\).
So, regardless of whether we are plugging in a positive or negative number in for \(x\) in the above, it seems that the result is exactly what you plugged in! If \(p\) is positive, and you plug \(x=p\) into \(\sqrt[3]{x^3}\), then you get out \(p\). If you plug in \(-p\), you get out \(-p\). Hence, for odd powers cancelling out odd roots, you don’t need absolute values around your solution (i.e. you don’t need to force the result to be nonnegative).
Radicals are Fractional Exponents!
For instance, here is how a few different radicals can be converted into their exponent counterpart:
$$\sqrt{x}=\sqrt[2]{x}=x^{\frac{1}{2}}$$
$$\sqrt[3]{x}=x^\frac{1}{3}$$
Using the Principle on More Complicated (Exponential) Expressions
The above principle/key idea gives us the ability to rewrite more complicated expressions, such as radicals of exponential expressions, more compactly. For instance, we could rewrite \(\sqrt[5]{x^3}\) as an exponential expression with a fractional exponent by noticing
$$\sqrt[5]{x^3}=(x^3)^{\frac{1}{5}}=x^\frac{3}{5}$$
Similarly, we can rewrite the expression \(\sqrt{(x+1)^5}\) as
$$\begin{align}\sqrt{(x+1)^6}&=((x+1)^5)^\frac{1}{2}\\&=(x+1)^\frac{5}{2}\end{align}$$
Swapping Radicals and Powers
Knowing that radicals are fractional exponents gives us a rather powerful ability: the ability to swap a radical with a power.
That is, if you are taking a root of the power of some expression, you can swap the power and the radical! i.e. if \(m\) is any real number and \(n\) is a natural number,
$$\sqrt[n]{x^m}=(\sqrt[n]{x})^m$$
That is, the power gets moved to the outside and the radical gets moved to the inside when you have a radical of a power. Similarly, if you reverse the equality, reading it right to left instead, the power gets moved to the inside and the radical gets moved to the outside when you’re taking the power of a radical.
Which one you use (i.e. radical on the inside or on the outside) is entirely dependent on what you are trying to achieve in a given problem.
Let \(m\) be any real number and suppose \(n\) is a positive whole number. Then, because radicals are just fractional powers, we notice that
$$\begin{align}\sqrt[n]{x^m}&=(x^m)^\frac{1}{n}\\&=x^{m\cdot \frac{1}{n}}\\&=x^{\frac{1}{n}\cdot m}\\&=(x^\frac{1}{n})^m\\&=(\sqrt[n]{x})^m\end{align}$$
This all works out because you can swap the order of the exponents being multiplied, as in step 3 above.
The ability to swap radicals with powers is useful in many contexts. For instance, sometimes it is easier to take roots of a quantity before raising them to a power, rather than the other way around. The following examples validate this.
In this case, if working this problem by hand (i.e. without a calculator) you might notice that it is ugly to square the \(27\) first and then take cube roots. i.e.
$$\sqrt[3]{27^2}=\sqrt[3]{729}=???$$
However, if we swap the radical and the power, things become much easier
$$\begin{align}\sqrt[3]{27^2}&=(\sqrt[3]{27})^2\\&=(3)^2\\&=9\end{align}$$
Much easier and tidier.
Solution: \(64\)
Note that since we are taking the root of a power, we can write
$$\begin{align}\sqrt{16^3}&=(\sqrt{16})^3\\&=4^3\\&=64\end{align}$$