Why are the graph of $f(x) = (x^2)^\frac{1}{6}$ and the graph of $f(x) = x^\frac{1}{3}$ not the same?

Why is the graph of $$f(x) = (x^2)^\frac{1}{6}$$ not the same graph as $$f(x) = x^\frac{1}{3}$$?

Shouldn't they be the same because when you apply the exponent rules to the first equation, you get the same result as the second equation?

By the way, I am an Algebra 2 student in High School.

• Hint-When you square something the negative disappears. – KingLogic Nov 25 '20 at 2:43
• Exponent rules don't work with negative bases and fractional exponents; cf. this question – J. W. Tanner Nov 25 '20 at 2:48
• The short answer is that $(x^p)^q=x^{pq}$ only for $x>0$ when $p$ and $q$ are rational numbers. – Clayton Nov 25 '20 at 2:50
• See also math.stackexchange.com/questions/2085268/does-i4-equal-1/…, and in particular the second-to-last paragraph. – mweiss Nov 25 '20 at 3:00
• Genuine question: is $-1^ \frac13$ where the operation " ^ $^ \frac13$ " is acting upon real numbers only and not in the complex realm, generally considered to be equal to $-1$, or is it undefined? Or do different texts have different conventions? Ah, J. W. Tanner has linked to the answer to my question... – Adam Rubinson Nov 25 '20 at 3:07

When you square the quantity $$x$$, the negative sign disappears, and it doesn't come back if you raise it to the $$\frac{1}{6}$$ power. If you just take $$\frac{1}{3}$$ power, then the negative stays negative.
$$((-1)^2)^\frac{1}{6}=1^\frac{1}{6}=1$$, while $$(-1)^\frac{1}{3}=-1$$
This hyperlink contains a graph of the two equations. Note that when $$x<0$$ the red solid line is the blue dotted line reflected about the x-axis.