TRICKY ONE

$\begin{array}{rcl} x+\frac1x&=&-1\\ x^2+1&=&-x\\ x^2+x+1&=&0&\mbox{---(1)}\\ (x-1)(x^2+x+1)&=&0\\ x^3-1&=&0\\ x^3&=&1&\mbox{---(2)} \end{array}$ $\begin{array}{cl} &x^7+x^2+1\\ =&x^3x^3x+x^2+1\\ =&x+x^2+1&\mbox{from (2)}\\ =&x^2+x+1\\ =&\fbox0&\mbox{from (1)} \end{array}$

it can be done using complex cube root of unity. x^2+x+1=0.put x=w (where w^3=1) therefore w^7 +w^2 +1=w^1+w^2 +1 and w^1+w^2 +1=0 .therefore answer is 0.

This is an easier solution to the problem: x + 1/x = -1 Thus, x^2 +1 = -x. Thus, required = x^7 + x^2 -1 = x(x^6 - 1). Now, x^2 +1 = -x x^2 +x +1 = 0 Multiply both sides by (x-1). (x-1)(x^2 +x +1) = 0 x^3 - 1 =0 x^3 =1 x^6 =1

Thus, required = x^7 + x^2 -1 = x(x^6 - 1) = 0.

x+ 1/x=-1. so x^2 + x +1=0...divide (x^7 + x^2 +1) by (x^2 +x+1) using long division...so we get quotient=x^5 - x^4 + x^2 -x+1 and remainder 0...Using divident=(divisor * quotient)+remainder, we get x^7 + x^2 +1=(x^2 + x+1)*(x^5 - x^4 + x^2 -x+1) + 0....since x^2 + x +1=0 therefore x^7 + x^2 +1=0...

1 st thing is u have to factorize it ,,,,( x^7-x^4)+(x^4+x^2+1),,,,then take the common term (x^2+x+1),, whose value is 0 ,,,as stated in the question ,,,,so the answer is zero x^4(x^3-1) + (x^4+2 x^2+1-x^2) {x^4(x-1)(x^2+x+1)} + {(x^2+x+1)(x^2-x+1)} (x^2+x+1) {x^4(x-1)+(x^2-x+1)} = 000