1.Let α\alphaα be a solution of x2+x+1=0x^{2} + x + 1 = 0x2+x+1=0, and for some a,b∈Ra, b \in \mathbb{R}a,b∈R, [4 a b](11613−1−12−2−14−8)=[0 0 0][4 \; a \; b] \begin{pmatrix} 1 & 16 & 13 \\ -1 & -1 & 2 \\ -2 & -14 & -8 \end{pmatrix} = [0 \; 0 \; 0][4ab]1−1−216−1−14132−8=[000]. If 4α4+mαa+nαb=34\alpha^{4} + m\alpha^{a} + n\alpha^{b} = 34α4+mαa+nαb=3, then m+nm + nm+n is equal to:a.333b.111111c.777d.888Login to continueOnly logged in users canattempt or see the solution.
