Prove that if `alpha, beta, gamma !=0` then `|(alpha+a_1b_1, a_1b_2, a_1b_3), (a_2b_1, beta+a_2b_2, a_2b_3), (a_3b_1, a_3b_2, gamma+a_3b_3)|=alpha beta gamma [1+(a_1b_1)/alpha + (a_2b_2)/beta+(a_3b_3)/gamma]`

|[2a_1b_1, a_1b_2+a_2b_1, a_1b_3+a_3b_1] , [a_1b_2+a_2b_1, 2a_2b_2, a_2b_3+a_3b_2] , [a_1b_3+a_3b_1, a_3b_2+a_2b_3, 2a_3b_3]|=

Show that if x_1, x_2, x_3!=0 |x_1+a_1b1a_1b_2a_1b_3a_2b_1x_2+a_2b_2a_2b_3a_3b_1a_3b_2x_3+a_3b_3|=x_1x_2x_3(1+(a_1b_1)/(x_1)+(a_2b_2)/(x_2)+(a_3b_3)/(x_3)) .

Show that if x_1, x_2, x_3!=0 |x_1+a_1b1a_1b_2a_1b_3a_2b_1x_2+a_2b_2a_2b_3a_3b_1a_3b_2x_3+a_3b_3|=x_1x_2x_3(1+(a_1b_1)/(x_1)+(a_2b_2)/(x_2)+(a_3b_3)/(x_3)) .

Show that if x_1, x_2, x_3!=0 |x_1+a_1b1a_1b_2a_1b_3a_2b_1x_2+a_2b_2a_2b_3a_3b_1a_3b_2x_3+a_3b_3|=x_1x_2x_3(1+(a_1b_1)/(x_1)+(a_2b_2)/(x_2)+(a_3b_3)/(x_3)) .

Show that if x_1, x_2, x_3!=0 |x_1+a_1b1a_1b_2a_1b_3a_2b_1x_2+a_2b_2a_2b_3a_3b_1a_3b_2x_3+a_3b_3|=x_1x_2x_3(1+(a_1b_1)/(x_1)+(a_2b_2)/(x_2)+(a_3b_3)/(x_3)) .

Prove that |[a_1alpha_1+b_1beta_1, a_1alpha_2+b_1beta_2, a_1alpha_3+b_1beta_3], [a_2alpha_1+b_2beta_1, a_2alpha_2+b_2beta_2, a_2alpha_3+b_2beta_3], [a_3alpha_1+b_3beta_1, a_3alpha_2+b_3beta_2, a_3alpha_3+b_3beta_3]|=0

Prove that |[a_1alpha_1+b_1beta_1, a_1alpha_2+b_1beta_2, a_1alpha_3+b_1beta_3], [a_2alpha_1+b_2beta_1, a_2alpha_2+b_2beta_2, a_2alpha_3+b_2beta_3], [a_3alpha_1+b_3beta_1, a_3alpha_2+b_3beta_2, a_3alpha_3+b_3beta_3]|=0

Show that | (1,1,1), (alpha ^ 2, beta ^ 2, gamma ^ 2), (alpha ^ 3, beta ^ 3, gamma ^ 3) | = (alpha-beta) (beta-gamma) (gamma-alpha) (alphabeta + betagamma + gammaalpha) |

Prove that |a_1alpha_1+b_1beta_1a_1alpha_2+b_2beta_2a_1alpha_3+b_1beta_3a_2alpha_1+b_2beta_1a_2alpha_2+b_2beta_2a_2alpha_3+b_2beta_3a_ 3alpha_1+b_3beta_1a_3alpha _2+b_3beta_2a_3alpha_3+b_3beta_3|=0