Find the remainder when x^2007 is divisible by (x^2-1)

Here the remainder is assumed to be a linear since the divisor is a quadratic equation.
Therefore the remainder is in the form of=$ax+b$
So,$x^{2007}$=$Quotient*(x^{2}-1)+(ax+b)$
By hit and trail method,
Putting the value of x as 1,we see $1=0+(a+b)$ and
Putting the value of x as -1,we see $-1^{2007}=-1=0+(-a+b)$
Thus By simultaneously solving the above equations we get the value of b as 0 and the value of a as 1
$\therefore$ Remainder=x+0=x

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