If xn xn+2 xn+3yn yn+2 yn+3zn zn+2 zn+3=(x−y)(y−z)(z−x)1x+1y+1z, then n equals
1
-1
2
-2
The degree of each term of the determinant is n + (n + 2) + (n + 3) = 3n + 5 and the degree of each term of the expression on R.H.S. is 2.
∴ 3n+5=2 or n=−1