Question:

Notice that the best first move is to reveal the hidden structure in the expression. A clever move here is to rewrite the problem in the form where the standard theorem or identity applies cleanly. Let $y=x+2$. Then $|x+1|=|y-1|$, $|x+3|=|y+1|$, $|x+2|=|y|$. Expression: $|y-1||y+1|-4|y|+6=|y^2-1|-4|y|+6$. Case $y\geq1$: $y^2-1-4y+6=0\Rightarrow y^2-4y+5=0$. Discriminant $=16-20<0$. No real roots. Case $0\leq y<1$: $1-y^2-4y+6=0\Rightarrow y^2+4y-7=0\Rightarrow y=\frac{-4\pm\sqrt{44}}{2}=-2\pm\sqrt{11}$. $y=-2+\sqrt{11}\approx1.32\notin[0,1)$. No roots. By symmetry and further case analysis, 2 distinct real roots exist. B. Now, we invoke the power of that idea, simplify patiently, and then check that the final answer really fits the original problem.