Show that #|(a+b,b+c,c+a),(b+c,c+a,a+b),(c+a,a+b,b+c)|=-2(a^3+b^3+c^3-3abc#?

Question

Please mention full procedure

877 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-08-06T18:22:18

Please see below.

We know that ,
#x^3+y^3+z^3#=#(x+y+z)(x^2#+#y^2#+#z^2-xy-yz-zx)+3xyz#
Let ,
#D=|(a+b,b+c,c+a),(b+c,c+a,a+b),(c+a,a+b,b+c)|#
Taking ,#R_1+R_2+R_3#

#D=|(2a+2b+2c,2a+2b+2c,2a+2b+2c),(b+c,c+a,a+b),(c+a,a+b,b+c)|#

Taking #R_1(1/(2a+2b+2c))#

#D=(2a+2b+2c)|(1,1,1),(b+c,c+a,a+b),(c+a,a+b,b+c)|#

Use , #C_2-C_1 and C_3-C_2#

#D=2(a+b+c)|(1,1-1,1-1),(b+c,c+a-b-c,a+b-c-a),(c+a,a+b-c-a,b+c-a-b)|#

#:.D=2(a+b+c)|(1,0,0),(b+c,a-b,b-c),(c+a,b-c,c-a)|#

Expanding we get

#D=2(a+b+c){1(a-b)(c-a)-(b-c)(b-c)-0=0}#

#:.D=2(a+b+c){ac-a^2-bc+ab-b^2+bc+bc-c^2}#

#:.D=2(a+b+c)(-a^2-b^2-c^2+ab+bc+ca)#

#:.D=-2(a+b+c)(a^2+b^2+c^2-ab-bc-ca)#

Using above formula for #x^3+y^3+z^3#

#:.D=-2(a^3+b^3+c^3-3abc)#