 Hello and welcome to the session. In this session we discussed the following question that says factorize x, y, z plus x complement, y complement, z complement. Let's now proceed with the solution. We are supposed to factorize x, y, z plus x complement, y complement, z complement. Now for this Boolean algebra, we have the elements a, b and c belonging to the set b. Then according to the distributive law we have a plus b into cd whole is equal to a plus bd whole into a plus cd whole. So applying this distributive law here we have this is equal to x, y, z plus x complement b whole into x, y, z plus y complement into z complement b whole. That is this expression is taken as a, this x complement as b and y complement, z complement as c and then the distributive law is applied. So further we get x, y, z plus x complement the whole into, now again we will apply the distributive law for this expression. So we get x, y, z plus y complement the whole into x, y, z plus z complement the whole. The sum operation is commutative. So we have a plus b is equal to b plus a, this is the commutative law. So applying this commutative law to the three expressions here we have this is equal to x complement plus x, y, z the whole into y complement plus x, y, z the whole into z complement. Now again we will apply the distributive law in these three terms separately. So when we apply the distributive law for this term x complement plus x, y, z we consider x complement as a then x as b and y, z as c. So on applying the distributive law we get this is equal to x complement plus x the whole into x complement plus y, z the whole. This into now again applying the distributive law to this term taking y complement as a, x as b, y, z as c we get y complement plus x the whole into y complement plus y, z the whole. This whole into in the same way applying the distributive law for this we get z complement plus x the whole into z complement plus y, z the whole. For any element a we have its inverse a complement such that a plus a complement is equal to 1 which is same as a complement plus a where this one is the identity element for the operation of the product. So x complement plus x could be written as 1 so this is equal to 1 into now x complement plus y, z could be written as x complement plus y the whole into x complement plus z that is we have applied the distributive law again. So this whole into y complement plus x the whole into applying the distributive law for this we get y complement plus y the whole into y complement plus z the whole. This whole into z complement plus x the whole into applying the distributive law for this we have z complement plus y the whole into z complement plus z the whole. Now as a plus a complement is equal to 1 which is same as a complement plus a so y complement plus y could be written as 1, z complement plus z could be written as 1. So this is equal to 1 into x complement plus y the whole into x complement plus z the whole into y complement plus x the whole into 1 into y complement plus z the whole into z complement plus z the whole into z complement plus z the whole into z complement plus z the whole. the whole into 1 the whole. Now we have A into 1 is equal to 1 into A is equal to A where this 1 is the identity element for the operation of the product. So using this we get this is further equal to x complement plus y the whole into x complement plus z the whole into y complement plus x the whole into y complement plus z the whole into z complement plus x the whole into z complement plus y the whole. So these are the factors of the given expression therefore we can say that x y z plus x complement y complement z complement can be factorized as x complement plus y the whole into x complement plus z the whole into y complement plus x the whole into y complement plus z the whole into z complement plus x the whole into z complement plus y the whole. So this is our final answer for each session hope you understood the solution of this question.