Given: x+1x=−2x+\frac{1}{x}=-2x+x1=−2, then determine the value of x6+1x6−3(x4+1x4)+4(x2+1x2)x^6+\frac{1}{x^6}-3\left(x^4+\frac{1}{x^4} \right)+4\left(x^2+\frac{1}{x^2} \right)x6+x61−3(x4+x41)+4(x2+x21) A5 B8 C3 D4 Show answer & explanationCorrect answer: DExplanationThe correct answer is option 4 i.e. 4. x+1x=−2x+\frac{1}{x}=-2x+x1=−2 ⇒ (x+1x)2=(−2)2=4\left(x+\frac{1}{x} \right)^2=(-2)^2=4(x+x1)2=(−2)2=4 ⇒ x2+1x2=4−2=2x^2+\frac{1}{x^2}=4-2=2x2+x21=4−2=2 (x4+1x4)=(x2+1x2)2−2=22−2=2\left(x^4+\frac{1}{x^4} \right)=\left(x^2+\frac{1}{x^2} \right)^2-2=2^2-2=2(x4+x41)=(x2+x21)2−2=22−2=2 (x6+1x6)=(x2+1x2)3−3(x2+1x2)=23−3(2)=8−6=2\left(x^6+\frac{1}{x^6} \right)=\left(x^2+\frac{1}{x^2} \right)^3-3\left(x^2+\frac{1}{x^2} \right)=2^3-3(2)=8-6=2(x6+x61)=(x2+x21)3−3(x2+x21)=23−3(2)=8−6=2 x6+1x6−3(x4+1x4)+4(x2+1x2)x^6+\frac{1}{x^6}-3\left(x^4+\frac{1}{x^4} \right)+4\left(x^2+\frac{1}{x^2} \right)x6+x61−3(x4+x41)+4(x2+x21) = 2 - 3(2) + 4(2) = 2 - 6 + 8 = 4