Question

use factors theirm to prove that x+a is a factor of (x^n+ a^n) for any odd positive integer n

Answer 1

whenever we put any negative number to odd power,we get that power of number with negative sign,like
$( - {1)}^{2} =( { - 1)}^{4} = 1 \: \: \: even \: power \\ ( - {1)}^{3} = ( { - 1)}^{5} = - 1 \: \: \: odd \: powers$
so,
$x + a = 0 \\ x = - a$
we put this factor into given polynomial to prove that x+a is factor
$( {x}^{n} + {a}^{n} ) \\ = (( { - a)}^{n} + {a}^{n} ) \\ = (- {a}^{n} + {a}^{n} ) \: \: \: \: because \: \: n \: is \: odd \\ = 0$
so x+a is a factor.