| 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 |
A) 10
B) 16
C) 6
D) 8
Correct Answer: C
Solution :
| \[{{(a+b)}^{2}}+{{(b+c)}^{2}}+{{(c+a)}^{2}}\] |
| \[={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{b}^{2}}+{{a}^{2}}+{{c}^{2}}+2\,(ab+bc+ca)\] |
| \[=2\,({{a}^{2}}+{{b}^{2}}+{{c}^{2}})+2\,(ab+bc+ca)\] |
| \[=2\,({{a}^{2}}+{{b}^{2}}+{{c}^{2}})+2\times 1\] |
| Now, \[(a+b+c)=2\] |
| On squaring both sides, we get |
| \[{{(a+b+c)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2\,(ab+bc+ca)\] |
| \[\Rightarrow \] \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}={{(2)}^{2}}-2\times 1=4-2=2\] |
| \[\therefore \] \[{{(a+b)}^{2}}+{{(b+c)}^{2}}+{{(c+a)}^{2}}\] |
| \[=2\times 2+2=4+2=6\] |
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