| Given that \[a+b+c=2\] and \[ab+bc+ca=1,\] then the value of \[{{(a+b)}^{2}}+{{(b+c)}^{2}}+{{(c+a)}^{2}},\] is [SSC (CGL) 2014] |
A) 10
B) 16
C) 6
D) 8
Correct Answer: C
Solution :
| Given, \[(a+b+c)=2\]and \[ab+bc+ca=1\] |
| Now, \[{{(a+b)}^{2}}+{{(b+c)}^{2}}+{{(c+a)}^{2}}\] |
| \[={{a}^{2}}+{{b}^{2}}+2ab+{{b}^{2}}+{{c}^{2}}+2bc+{{c}^{2}}+{{a}^{2}}+2ca\] |
| \[={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ca+{{a}^{2}}+{{b}^{2}}+{{c}^{2}}\] |
| \[={{(a+b+c)}^{2}}{{a}^{2}}+{{b}^{2}}+{{c}^{2}}\] |
| \[={{(a+b+c)}^{2}}+{{(a+b+c)}^{2}}-2ab-2bc-2ca\] |
| \[=2\,\,{{(a+b+c)}^{2}}-2\,\,(ab+bc+ca)\] |
| \[=2\times \,\,{{(2)}^{2}}-2\times (1)=8-2=6\] |
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