Prove that `Axx(BcupC)=(AxxB)cup(AxxC)`

Let A={x in N:x^2-5x+6=0}. B={x in W:0lex<2} and C={x in N:x<3} ,then verify that Axx(BcupC)=(AxxB)cup(AxxC)

Let A={x""inNN:x^(2)-5x+6=0},B={x""inW:0lexlt2}andC={x""inNN:xlt3} , then verify that Axx(BcupC)=(AxxB)cup(AxxC)

If A={1,4},B={4,3},andC={3,6} , show that, A xx(BcupC)=(AxxB)cup (AxxC)

If A = {1, 3}, B = {3, 5} and C = {5, 10}, show that, Axx(Bcup C) =(AxxB) cup(AxxC)

If A={a,b},B={m,n}andC = {p,q}, show that, Axx(BcapC) =(AxxB)cap(AxxC)

If A = {1, 3}, B = {3, 5} and C = {5, 10}, show that, Axx(BcapC) =(AxxB) cap(AxxC)

For any three sets A,B and C prove that, Axx(B-C)=(AxxB)-(AxxC) .

For any three sets A, B, C, prove that A-(BcupC)=(A-B)cap(A-C)

For any three sets A,B and C, prove that, A-(BcupC)=(A-B) cap(A-C)

For any three sets A,B and C ,prove that A-(BcupC)=(A-B)cap(A-C) .