How to write this boolean expression using only NOR gates?
I'll provide some direction to head. Since your question isn't so much about simplification, I'll do the simplification without any further ado. Then I'll provide a direction to head from that point but I'll leave some work you'll need to perform yourself.

Simplify

Let's simplify the expression:

$$\begin{align*}
T &amp;= A\,\overline{B}\,C + \overline{A}\,B\,\overline{C} + A\,B\\
 &amp;= A\,\overline{B}\,C + \overline{A}\,B\,\overline{C} + A\,B\,\overline{C} + A\,B\,C\\
 &amp;= A\left(\overline{B}\,C + B\,\overline{C} + B\,C\right) + \overline{A}\,B\,\overline{C}\\
 &amp;= A\,\overline{\overline{B}\,\overline{C}} + \overline{A}\,B\,\overline{C}\\
 &amp;= A\,B + A\,C + \overline{A}\,B\,\overline{C}\\
 &amp;= A\,C + B\left(A + \overline{A}\,\overline{C}\right)\\
 &amp;= A\,C + B\left(A + \overline{C}\right)\\
 &amp;= A\,C + A\,B + B\,\overline{C}
\end{align*}$$

I'm not going to explain all the details above. It isn't the only, nor even the best approach to simplification. It's just one I decided to write out, off-hand. Still, you should try to take the time to understand each transition on your own. (Or not. It's up to you.)

From there, you can see that if \$A\$ and \$B\$ are both true, the expression \$T=A\,C+B\,\overline{C}\$ already captures the term, \$A\,B\$, as one or the other is picked up regardless of \$C\$.

So the simplified version is:

$$T=A\,C+B\,\overline{C}$$

Applying the NOR gate template

The basic model of a NOR gate, as I'm sure you know, is \$T=\overline{R+S}\$. That's the template. The question is, how do you take an arbitrary expression and make it conform?

I'll start you out:

$$\begin{align*}
T&amp;=A\,C+B\,\overline{C}\\\\
&amp;=\overline{\overline{A\,C+B\,\overline{C}}}\\\\
&amp;=\overline{\overline{A\,C}\cdot\overline{B\,\overline{C}}}\\\\
&amp;=\overline{\left(\overline{A}+\overline{C}\right)\cdot\left(\overline{B}+C\right)}\\\\
&amp;=\overline{\overline{A}\,\overline{B}+\overline{A}\,C+\overline{B}\,\overline{C}}\\\\
&amp;=\overline{\overline{A}\left(\overline{B}+C\right)+\overline{B}\,\overline{C}}
\end{align*}$$

(If it's not immediately clear, take note that in the first transition I used a double negation. The reason is that I know the basic NOR template requires a negation of the entire underlying expression and that a very obvious and simple way to achieve that is to double-negate the entire expression. This guarantees a negation that I can keep at the top, while applying the remaining negation to the underlying expression.)

At this point, you can see that if \$T=\overline{R+S}\$ then \$R=\overline{A}\left(\overline{B}+C\right)\$ and \$S=\overline{B}\,\overline{C}\$. So you are closer to an answer, now. But you have two new situations to resolve.

I'll solve \$S\$:

$$\begin{align*}
S&amp;=\overline{B}\,\overline{C}\\\\
&amp;=\overline{\overline{\overline{B}\,\overline{C}}}\\\\
&amp;=\overline{B+C}\\\\
\end{align*}$$

And that easily fits the model of a NOR with no additional work.

You get to resolve \$R\$, now. Once you walk yourself through that process, you should find a requirement for exactly five NOR gates. I've only shown you the first two of them. You get to find the other three.