Always true?

Question

is (x^3+a^3) always precisely divisible by (x+a)? What's the proof?

Thanks

Answer 1

It is not divisible by x+a if x+a=0, because nothing is divisble by zero (not even zero).

Answer 2

x^3+a^3 = (x+a) ( x^2-ax + a^2)

(x+a) ( x^2-ax + a^2) = x^3 - ax^2 + a^2x + ax^2 - a^2x + a^3
. . . . . . . . . . .. = x^3+a^3

(x+a) . . .. is a factor

Answer 3

Yes.
x³+a³ = (x+a)(x²-ax+a²).