Boolean algebra simplification problem

Question

I can't solve this equation:

$$(xy + x'yz)(xz + x'y') = xyz$$

After applying distribution I got this:

$$xyz + yz + x'z = xyz$$

I can't find the answer and have been thinking for hours now.

Answer 1

You should get four terms by distributivity:

\begin{align} (xy \vee \bar xyz)(xz \vee \bar x \bar y) &= xyxz \vee xy\bar x\bar y \vee \bar xyz xz \vee \bar xyz\bar x\bar y. \end{align}

Except from $xyz$ every term contains a variable and its negation, which results in $0$.

Answer 2

$(xy+x'yz)(xz+x'y')=xxyz+xx'yy'+xx'yzz+x'x'yy'z=xyz+0 \times 0+0 \times yz + x' \times 0 z = xyz$

$\square$