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) ne 0 |{:(x_(1) +a_(1)b_(1),,a_(1)b_(2),,a_(1)b_(3)),(a_(2)b_(1),,x_(2)+a_(2)b_(2),,a_(2)b_(3)),(a_(3)b_(1),,a_(3)b_(2),,x_(3)+a_(3)b_(3)):}| =x_(1)x_(2)x_(3) (1+(a_(1)b_(1))/(x_(1))+(a_(2)b_(2))/(x_(2))+(a_(3)b_(3))/(x_(3)))

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]

|(a_1-b_1+x,a_1-b_2,a_1-b_3),(a_2-b_1,a_2-b_2+x,a_2-b_3),(a_3-b_1,a_3-b_2,a_3-b_3+x)| =x^3+x^2sum_(i=l)^3(a_i-b_i)+xsum_(l lt=i < j <= 3)^3 (a_i-b_i)+x sum_(l <= i < j <= 3)(a_i-b_j)(b_i-b_j).

Find the coefficient of x in the determinant |((1+x)^(a_1 b_1),(1+x)^(a_1 b_2),(1+x)^(a_1 b_3)),((1+x)^(a_2 b_2),(1+x)^(a_2 b_2),(1+x)^(a_2 b_3)),((1+x)^(a_3 b_3),(1+x)^(a_3 b_2),(1+x)^(a_3 b_3))|=0.

If a_i, b_i in N for i 1,2,3, then coefficient of x in the determinant;|((1+x)^(a_1b_1),(1+x)^(a_1b_2),(1+x)^(a_1b_3)),((1+x)^(a_2b_1),(1+x)^(a_2b_2),(1+x)^(a_2b_3)), ((1+x)^(a_3b_1),(1+x)^(a_3b_2),(1+x)^(a_3b_3))|

Prove by vector method that (a_1b_1+a_2b_2+a_3b_3)^2lt+(a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2)

If (3)/((x+2)(2x+1))=(a)/(2x+1)+(b)/(x+2) then b= ?