using property of stp prove that [a bar+b bar    b bar+c bar    c bar+a bar]=2[a bar b bar c bar]

Dear Student,
Please find below the solution to the asked query:

LHS=a+bb+cc+a=a+b·b+c×c+a=a+b·b×c+a+c×c+a=a+b·b×c+b×a+c×c+c×a=a+b·b×c+b×a+0+c×a=a+b·b×c+b×a+c×a=a·b×c+b×a+c×a+b·b×c+b×a+c×a=a·b×c+a·b×a+a·c×a+b·b×c+b·b×a+b·c×a=a·b×c+a·b×a+a·c×a+b·b×c+b·b×a+b·c×a=abc+0+0+0+0+bca=abc+abc=2abc=RHS

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