Boolean Algebra Question

Factor to obtain a product of sums. (three terms)
WXY + WX'Y + WYZ + XYZ'

Please show work.
Thanks in advance!

f = WXY + WX'Y + WYZ + XYZ'

Step 1: Use DeMorgan's law to find f'.

f' = (WXY + WX'Y + WYZ + XYZ')'
f' = (W' + X' + Y') (W' + X + Y') (W' + Y' + Z') (X' + Y' + Z)

Step 2: Multiply out to put this in sum of products form.
f' = (W'W' + W'X + W'Y' + X'W' + X'X + X'Y' + Y'W' + Y'X + Y'Y') (W'X' + W'Y' + W'Z + Y'X' + Y'Y' + Y'Z + Z'X' + Z'Y' + Z'Z)
Step 3: Simplify along way.
f' = (W' + W'X + W'Y' + X'W' + X'Y' + Y'W' + Y'X + Y') (W'X' + W'Y' + W'Z + Y'X' + Y' + Y'Z + Z'X' + Z'Y')
f' = (W' (1 + X + Y' + X' + Y') + Y' (X' + X + 1)) (W'X' + W'Y' + W'Z + Y' (X' + 1 + Z + Z') + Z'X')
f' = (W' + Y') (W'X' + W'Y' + W'Z + Y' + Z'X')
f' = W'W'X' + W'W'Y' + W'W'Z + W'Y' + W'Z'X' + Y'W'X' + Y'W'Y' + Y'W'Z + Y'Y' + Y'Z'X'
f' = W'X' + W'Y' + W'Z + W'Y' + W'Z'X' + Y'W'X' + W'Y' + Y'W'Z + Y' + Y'Z'X'
f' = Y' (1 + W' + W' + W'X' + W' + W'Z + Z'X') + W'X' (1 + Z') + W'Z
f' = Y' + W'X' + W'Z

Step 4: Use DeMorgan's law to find f as a product of sums.
f = (Y' + W'X' + W'Z)'
f = Y (W + X) (W + Z')