Question

Is $x^7+a^7$divisible by $x+a$?

Answer 1

Yes, it is divisible, as there is no remainder.

Following are the steps for division:

Step $1.$ Multiply $x$ with $x^6$ to get $x^7$, Remainder $=-a{x}^6+a^7$

Step $2.$ Multiply $x$ with $-a{x}^5$to get $-a{x}^6$, remainder= $a^2{x}^5+a^7$

Our target would be to reduce the first term of the remainder.

This steps will continue to go unless all the terms are cancelled by division.

Step 6: Multiply $x$ with $-a^{5}x$ to get$-a^5{x}^2$, Remainder $=a^{6}x+a^7$

Step 7: Multiply $x$ with $a^6$ to get $a^{6}x$, Remainder $=0$