[10000印刷√] 2^x=3^y=12^z then xy= 130521-2^x=3^y=12^z then xy=

 So substituting the values for the elementary symmetric polynomials that we found, we find that x, y and z are the three roots of t3 −t2 − 1 2t − 1 6 = 0 or if you prefer 6t3 − 6t2 −3t −1