if a+b+c=1,a^2+b^2+c^2=21and ABC=8 then find the value of (1-a)(1-b)(1-c).
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(1-a) (1-b)(1-c) =(1-b-a+ab) (1-c)
=1-c-b+bc-a+ac+ab-abc
=1-(a+b+c) +(ab+bc+ac) -abc
=1-1+(ab+bc+ac) -8
=(ab+bc+ac) -8.....................(1)
Now
(a+b+c) ²=a²+b²+c²+2ab+2bc+2ca OR 2(ab+bc+ca) =(a+b+c) ²-(a²+b²+c²)
=(1)²-21=1-21= -20
ab+bc+ca= -10...................(2)
Now from (1) and (2) we get
(1-a) (1-b) (1-c) = -10-8=-18
=1-c-b+bc-a+ac+ab-abc
=1-(a+b+c) +(ab+bc+ac) -abc
=1-1+(ab+bc+ac) -8
=(ab+bc+ac) -8.....................(1)
Now
(a+b+c) ²=a²+b²+c²+2ab+2bc+2ca OR 2(ab+bc+ca) =(a+b+c) ²-(a²+b²+c²)
=(1)²-21=1-21= -20
ab+bc+ca= -10...................(2)
Now from (1) and (2) we get
(1-a) (1-b) (1-c) = -10-8=-18
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