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?

Please mention full procedure

1 Answer
maganbhai P.
Aug 6, 2018

Please see below.

Explanation:

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)

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