Remainder

Write $P(x)$ as $P(x)=x^9 - x^6 + x^3 - 1 = (x^3 - 1)(x^6 + 1) = (x - 1)(x^2 + x + 1)(x^6 + 1)$ So, $P$ is divisible by $d(x) = x^2 + x + 1$ , and therefore $r(x) = 0$ . Thus $r(2015) = \boxed{0}$ .

I divided straight away and found that x^9 - x^6 + x^3 - 1 = (x^2 + x + 1)(x^7 - x^6 + x - 1);

The remainder is a constant 0;

Therefore, r (2015) = 0.

$x^{ 3 }-1=(x-1)({ x }^{ 2 }+x+1)$ . Thus, $x^{ 3 }-1$ leaves $0$ as remainder in the division by $({ x }^{ 2 }+x+1)$

Hence:

$x^{ 3 }-1\equiv 0(mod\quad { x }^{ 2 }+x+1)\quad \Longleftrightarrow \quad x^{ 3 }\equiv 1(mod\quad { x }^{ 2 }+x+1)$

So, by the properties of Modular Arithmetics:

${ ({ x }^{ 3 }) }^{ 3 }\equiv { (1) }^{ 3 }(mod\quad { x }^{ 2 }+x+1)\quad \Longleftrightarrow \quad x^{ 9 }\equiv 1(mod\quad { x }^{ 2 }+x+1)\quad \\ { -({ x }^{ 3 }) }^{ 2 }\equiv -{ (1) }^{ 2 }(mod\quad { x }^{ 2 }+x+1)\quad \Longleftrightarrow \quad -x^{ 6 }\equiv -1(mod\quad { x }^{ 2 }+x+1)\\ x^{ 3 }\equiv 1(mod\quad { x }^{ 2 }+x+1)\\ -1\equiv -1(mod\quad { x }^{ 2 }+x+1)$

Summing it up, we get:

${ x }^{ 9 }-{ x }^{ 6 }+{ x }^{ 3 }-1\equiv 1-1+1-1(mod\quad { x }^{ 2 }+{ x }+1)\Longleftrightarrow \boxed { p(x)\equiv 0(mod\quad { x }^{ 2 }+x+1) }$

So $r(x)=0$ and then $\boxed { r(2015)=0 }$

$x^9-x^6+x^3-1=(x^2+x+1)(x^7-x^6+x-1)+0$ so the answer is [0]

I think problem is a bit overrated

Given polynomial is divisible by the given divisor.so remainder is 0.