Question
Prove that $\left|\begin{array}{cc}x^2+x+1 & x+1 \\ x & x-1\end{array}\right| =x^3-x^2-x-1$
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Answer
Suppose that $\Delta=\left|\begin{array}{cc}x^2+x+1 & x+1 \\ x & x-1\end{array}\right|$
expanding the determinant
$\Delta=\left(x^2+x+1\right)(x-1)-(x+1) x$
$=x^3-1-x^2-x$
$=x^3-x^2-x-1 \text { Hence proved. }$
expanding the determinant
$\Delta=\left(x^2+x+1\right)(x-1)-(x+1) x$
$=x^3-1-x^2-x$
$=x^3-x^2-x-1 \text { Hence proved. }$
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