Not really. If you "square root" both sides, you get $|x|=7$, which means $x=7$ or $x=-7$. Alternatively, you can subtract $48$ to get $x^2-49=0$
$x^2-7^2=0$
$(x-7)(x+7)=0$
$x-7=0\text{ or } x+7=0$
The cube root has the "nice" property is is a one one correspondence, from where there is "no harm" if you "cube root" you equation. This means $x^3=y^3 \iff x=y$
That is, two real numbers have the same third power if and only if they are the same number in the first place.
The problem with the square root (and in general even powers of $x,y$) is that
$x^n=y^{n} \iff x=y$
is false since the additive inverse also works, for example $(-7)^2=7^2$ yet $7=-7$ is manifestly false.
Your equation is
$x^3-1=0$
If you're working over $\Bbb R$, that is, considering real numbers as solutions, then there is no harm on doing what you suggest
$x^3-1=0$
$x^3=1\iff x=1$
Note $-1$ is not a solution since $(-1)^3=-1$. Now, the problem arises when we extend our "workplace" to the complex number system. Then our useful assertion that $\tag 1 x^3=y^3\iff x=y$ becomes false. In such case, you must factor the polynomial, as you did, and find complex solutions, since using $(1)$ would be wrong, for it is a true statement for real numbers only. You have found the counterexamples just now $\left(\frac{-1+i\sqrt 3}{2}\right)^3=\left(\frac{-1-i\sqrt 3}{2}\right)^3=1^3=1$ but $\frac{{ - 1 + i\sqrt 3 }}{2} \ne \frac{{ - 1 - i\sqrt 3 }}{2} \ne 1$