x ²ⁿ – y²ⁿ is divisible by x + y.

Let P(n) = x ²ⁿ – y²ⁿ
P(1) = x² – y²
        = (x + y) (x-y) which is divisible by (x+y)
P(1) is true
Let us assume P(k) is true
P(k) = x^2k-y^2k = m(x+y)……………. A
TO Prove P(k+1) is divisible by (x+y) using the resultsof A
P(k +1) = x^2(k+1) – y^2(k+1)

P(k+1) = x^2(k+1) – y^2(k+1)
            = x^2k . x² – y^2k . y²
            = x^2k . x² – [ x^2k – m(x+y)]y² From result A
            = x^2k . x² – y².x² + my² (x+y)
            = x^2k (x² – y²) + my² (x+y)
y^2k = x^2k - m(n+y)
     = (x+y) { x^2k (x-y) + my²}
P(k+1) is divisible by (x+y)
hence the result.
P(K+1) is true.
By the Principle of mathematical induction, P(n) is true for all values of nwhere n N
Hence proved