Question

prove the following determinant

please answer me fast, ASAP​

Answer 1

Answer:

$l.h.s. = 1(a {b}^{2} - a {c}^{2} ) - 1( {a}^{2} b - b {c}^{2} ) + 1( {a}^{2} c - {b}^{2} c) \\ \\ = a {b}^{2} - a {c}^{2} - {a}^{2} b + b {c}^{2} + {a}^{2} c - {b}^{2} c\\ \\ = (a {b}^{2} - {a}^{2} b) + ( {a}^{2} c - {b}^{2} c) + (b {c}^{2} - a {c}^{2} ) \\ \\ = - ab(a - b) + c(a + b)(a - b) + - {c}^{2} (a - b) \\ \\ = (a - b)( - ab + ac + bc - {c}^{2} ) \\ \\ = (a - b)( - a(b - c) + c(b - c) \\ \\ = (a - b)(b - c)(c - a) \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: hence \: proved$

Answer 2

=1(ab2-ac2)-1(a2b-bc2)+1(a2c - ac)

= ab2 - ac2 -a2b +bc2 + a2c - b2c

= (ab2- a2b) + (a2c-b2c) +(bc2- ac2)

= -ab(a-b) +c (a+b)(a-b) - c2(a-b)

= (a-b) (-ab +ac+bc -c2)

= (a-b ) (-a ( b-c) + c(b- c)

= (a- b) ( b-c) ( c- a)