If is a non-real cube root of unity, then what is a root of the following equation? vmatrix x+1 & & ^2 & x+ ^2 & 1 ^2 & 1 & x+ vmatrix = 0

Step-by-step solution

Applying the column operation C₁ C₁ + C₂ + C₃ , the determinant becomes: vmatrix x+1+ + ^2 & & ^2 x+1+ + ^2 & x+ ^2 & 1 x+1+ + ^2 & 1 & x+ vmatrix = 0 Since 1 + + ^2 = 0 for a non-real cube root of unity, the equation simplifies to: vmatrix x & & ^2 x & x+ ^2 & 1 x & 1 & x+ vmatrix = 0 Taking x common from the first column: x vmatrix 1 & & ^2 1 & x+ ^2 & 1 1 & 1 & x+ vmatrix = 0 This implies x = 0 is a root of the equation. Answer: x = 0