<s>[INST] How to square both the sides of an equation? Question: $x^2  \sqrt{(x + 3)} = (x + 3)^{3/2}$
My solution: $x^4 (x + 3) = (x + 3)^3$
$=> (x + 3)^2 = x^4$
$=> (x + 3)  = x^2$
$=> x^2 -x - 3 = 0$
$=> x = (1 \pm \sqrt{1 + 12})/2$
I understand that you can't really square on both the sides like I did in the first step, however, if this is not the way to do it, then how can you really solve an equation like this one (in which there's a square root on the LHS) without substitution? 
 [/INST] Why can't you square both sides like you did?  You absolutely can.  If you do this you just need to make sure you didn't introduce any extraneous solutions.  You can make sure of this by checking each solution you get.  Admittedly that may be a little difficult with answers like $x = (1 \pm \sqrt{13})/2$, but it is what it is.
What I'm not sure of is how your $x^4$ became an $x^2$ when you divided both sides by $x+3$.  Actually now that I look closer that $x^2$ just looks like a typo.  Anyway.. I would generally not divide like that because then you lose solutions.  In this case you lost the solution $x = -3$.  It's better to factor rather than divide.
Recall that $y^{3/2} = y\sqrt{y}$, so in particular we have $(x+3)^{3/2} = (x+3)\sqrt{x+3}$.  If you want to do this without squaring both sides, I'd proceed like this:
\begin{align}
  x^2\sqrt{x+3} &= (x+3)^{3/2}\\[0.3cm]
  x^2\sqrt{x+3} - (x+3)^{3/2} &= 0\\[0.3cm]
  x^2\sqrt{x+3} - (x+3)\sqrt{x+3} &= 0\\[0.3cm]
  \sqrt{x+3}\left(x^2 - (x+3)\right) &= 0\\[0.3cm]
  \sqrt{x+3}(x^2-x-3) &= 0
\end{align}
So either $\sqrt{x+3} = 0$ or $x^2 - x - 3 = 0$.  The first gives $x = -3$ and the second gives $x = (1 \pm \sqrt{13})/2$.
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