Prove that `|b+c a a b c+a b cc a+b|=4a b c`

Prove that |b+c a a b c+a b c c a+b |

Prove: |b+c a a b c+a b cc a+b|=4a b c

Using the property of determinants and without expanding, prove that: |-a^2a b a c b a b^2b cc a c b-c^2|=4a^2b^2c^2

Prove that |b c-a^2c a-b^2a b-c^2-b c+c a+a bb c-c a+a bb c+c a-a b(a+b)(a+c)(b+c)(b+a)(c+a)(c+b)|=3.(b-c)(c-a)(a-b)(a+b+c)(a b+b c+c a)

Prove that =|a cc-a a+cc bb-c b+c a-bb-c0a-c x y z1+x+y|=0 implies that a ,b ,c are in A.P. or a ,c ,b are in G.P.

Prove that: |(-2a , a+b , a+c ) ( b+a ,-2b , b+c ) ( c+a, c+b , -2c)| =4(a+b)(b+c)(c+a)

Prove that |(b+c,a,a),(b,c+a,b),(c,c,a+b)|= 4ac

Prove that: |1a a^2-b c1bb^2-c a1cc^2-a b|=0

Prove that: |1a a^2-b c1bb^2-c a1cc^2-a b|=0

Prove that |a ,b+c, a^2,b, c+a, b^2,c, a+b, c^2|=-(a+b+c)xx(a-b)(b-c)(c- a)