Question

solve it step by step​

Answer 1

Answer:

(i) By the Factor Theorem, x-a is a factor of f(x) if f(a) = 0.

Putting f(x) = xⁿ-aⁿ gives f(a) = aⁿ-aⁿ = 0.

So x-a is a factor of xⁿ-aⁿ.

(ii) By the Factor Theorem, x+a = x-(-a) is a factor of f(x) if f(-a) = 0.

Putting f(x) = xⁿ+aⁿ gives f(-a) = (-a)ⁿ+aⁿ = -aⁿ+aⁿ (since n is odd) = 0.

So x+a is a factor of xⁿ+aⁿ when n is odd.