**Is the following proof valid? If not, what is the invalid step in the proof?**
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:- | :-
| $\phantom{=}(2x+1)^2-(x+1)^2$
Step $1\quad$ | $=4x^2+4x+1-x^2+2x+1$
Step $2$ | $=3x^2+6x+2$
Step $3$ | $=3x^2+6x+3-1$
Step $4$ | $=3(x^2+2x+1)-1$
Step $5$ | $=3(x+1)^2-1$
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