Is $x^{8} + a^{8}$ divisible by $x + a$ ?

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Solution

Step 1) Multiply $x + a$ with $x^{7}$ ,to get $x^{8}$ ,Remainder $= - a x^{7} + a^{8}$
Step 2) Multiply $x + a$ with $- a x^{6}$ ,to get $- a x^{7}$ , Remainder $= a^{2} x^{6} + a^{8}$
Continue it for further 5 steps.
Step 8) Multiply $x + a$ with $- a^{7}$ , to get, $- a^{7} x + a^{8}$ , Remainder $= 2 a^{8}$
$∵$ Remainder $= 2 a^{8}, x^{8} + a^{8}$ is not exactly divisible by $x + a$

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