[" Prove that "([3a,-a+b,-a+c],[-b+a,3b,-b+c],[-c+a,-c+b,3c])=3(a+b+c)(ab+bc+ca)]

Using properties of determinants, prove that: |[3a,-a+b,-a+c],[-b+a,3b,-b+c],[-c+a,-c+b,3c]| = 3(a+b+c)(ab+bc+ca)

Using properties of determinants, prove that |[3a,-a+b,-a+c],[-b+a,3b,-b+c],[-c+a,-c+b,3c]| = 3(a+b+c)(ab+bc+ca)

Show that [[3a,-a+b,-a+c],[-b+a,3b,-b+c],[-c+a,-c+b,3c]] = 3(a+b+c)(ab+bc+ca)

Show that |(3a,-a+b,-a+c),(-b+a,3b,-b+c),(-c+a,-c+b,3c)|=3(a+b+c)(ab+bc+ca)

Using properties of determinants in Exercise 11 to 15 prove that |{:(3a,-a+b,-a+c),(-b+a,3b,-b+c),(-c+a,-c+b,3c):}|=3(a+b+c)(ab+bc+ca)

Show that: ,3a,-a+b,-a+c-b+a,3b,-b+c-c+a,-c+b,3c]|=3(a+b+c)(ab+bc+ca)

Prove that |(3a, -a+b, -a+c),(a-b, 3b, c-b),(a-c, b-c, 3c)| = 3(a+b+c)(ab+bc+ca) .

3a,-a+b,-a+ca-b,3b,c-ba-b,3b,c-ba-c,b-c,3c]|=

{:|( 3a,-a+b,-a+c),( -b+a, 3b,-b+c) ,( -c+a,-c+b,3c) |:} = 3( a+b+c) ( ab+bc+ca)

Using the property of determinants prove that {:|( 3a,-a+b,-a+c),( -b+a, 3b,-b+c) ,( -c+a,-c+b,3c) |:} = 3( a+b+c) ( ab+bc+ca)