Prove that `x^m+y^m` is divisible by x + y, when m is an odd natural number.

Statement - 1 : For all n in N x^(n) - y^(n) is divisible by x-y Statement - 2: x^(n) + y^(n) is divisible by x+y if n is even natural number Which of the above statement is true :

Prove that x^n-y^n is divisible by x - y for all positive integers n.

Match the roster form with set builder form. (i) {1,2,3,6} (a) {x : x is prime number and a divisior of 6} (ii) {2,3} (b) {x : x is an odd natural number smaller than 10} (iii) {m,a,t,h,e,m,a,t,i,c,s} (c) {x : x is a natural number and divisor of 6} (iv) {1,3,5,7,9} (d) {x : x is a letter of the word MATHEMATICS}

If A = { x : x is a natural number }, B = { x : x is an even natural number} C = { x : x is an odd natural number} and D = { x : x is a prime number }, find B ∩ C

If A = { x : x is a natural number }, B = { x : x is an even natural number} C = { x : x is an odd natural number} and D = { x : x is a prime number }, find B ∩ D

If A = { x : x is a natural number }, B = { x : x is an even natural number} C = { x : x is an odd natural number} and D = { x : x is a prime number }, find A ∩ B

If A = { x : x is a natural number }, B = { x : x is an even natural number} C = { x : x is an odd natural number} and D = { x : x is a prime number }, find A ∩ C

If A = { x : x is a natural number }, B = { x : x is an even natural number} C = { x : x is an odd natural number} and D = { x : x is a prime number }, find A ∩ D

If A = { x : x is a natural number }, B = { x : x is an even natural number} C = { x : x is an odd natural number} and D = { x : x is a prime number }, find C ∩ D