Boolean Simplification Problem I have to simplify the figure

I came up with the equation 
$$(ac+\bar b)\cdot (a+b+\bar c) \cdot (b+ac).$$
Help me out 
Attempted Answer:
\begin{align*}
(ac+\bar b)\cdot (a+b+c)\cdot (b+ac)
&=(ac+abc+ac+a\bar b+\bar b c)\cdot(b+ac)\\
&=abc+abc+abc+ac+abc+ac+a\bar b c+a\bar b c \\
&=abc+a\bar b c+ac\\
&=ac(b+\bar b)+ac\\
&=ac+ac=ac
\end{align*}

OK, here goes:
We need to simplify
$(ac + \bar b)(a + b + \bar c)(b + ac). \tag{1}$
I'll do this by a method which is a little different than the OP's, but will yield the same answer $ac$; this will corroborate k31453's work.  First, I observe that 
$(ac + \bar b)(a + b + \bar c)(b + ac) = (ac + \bar b)(ac + b)(a + b + \bar c), \tag{2}$
and we have
$(ac + \bar b)(ac + b) = (ac)(ac) + acb + ac \bar b + \bar b b = ac + acb + ac \bar b, \tag{3}$
since $(ac)(ac) = ac$ and $\bar b b = 0$.  Next, I use
$acb + ac \bar b = ac(b + \bar b) = ac1 = ac \tag{4}$
to obtain 
$(ac + \bar b)(ac + b) = ac + ac = ac. \tag{5}$
Substituting this back into (2) yields
$ac(a + b + \bar c) = ac + abc = ac(1 + b) = ac1 = ac, \tag{6}$
confirming k31453's calculation.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
