Given that, x+ab+c+2a+x+bc+a+2b+x+ca+b+2c=0 =x+ab+c+2a-1+x+bc+a+2b-1+x+ca+b+2c-1=0 =x+a-b-c-2ab+c+2a+x+b-c-a-2bc+a+2b+x+c-a-b-2ca+b+2c=0 =x-a-b-cb+c+2a+x-a-b-cc+a+2b+x-a-b-ca+b+2c=0 =x-a-b-c(1b+c+2a+1c+a+2b+1a+b+2c) Now, (1b+c+2a+1c+a+2b+1a+b+2c)≠0 ∴x-a-b-c=0 =x=(a+b+c)
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