If x=ω−ω2−2 then the value of x4+3x3+2x2−11x−6 is (where ω is cube root of unity)______
We have x=ω−ω2−2 or x+2=ω−ω2Squaring,x2+4x+4=ω2+ω4−2ω3=ω2+ω−2=−3⇒x2+4x+7=0Dividing x4+3x3+2x2−11x−6 by x2+4x+7 we getx4+3x3+2x2−11x−6=x2+4x+7x2−x−1+1=(0)x2−x−1+1=0+1=1