a1× b1+ b1× c1+ c1× a1=

=b×c[a b c]×c×a[a b c]+c×a[a b c]×a×b [a b c]+a×b[a b c]×b×c[a b c] =[b c a] c-0[a b c]2+[c a b][a b c]2+[a b c] b-0[a b c]2 =a+b+c[a+b+c]

a1× b1+ b1× c1+ c1× a1=

=b×c[a b c]×c×a[a b c]+c×a[a b c]×a×b [a b c]+a×b[a b c]×b×c[a b c] =[b c a] c-0[a b c]2+[c a b][a b c]2+[a b c] b-0[a b c]2 =a+b+c[a+b+c]

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$a^{1} \times b^{1} + b^{1} \times c^{1} + c^{1} \times a^{1} =$

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$\frac{a + b + c}{[b a c]}$

$\frac{a + b + c}{2 [a b c]}$

$\frac{a + b + c}{[a b c]}$

$\frac{a - b - c}{[a b c]}$

answer is A.

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Detailed Solution

$\begin{array}{rcl} = & \frac{b \times c}{[a b c]} \times \frac{c \times a}{[a b c]} + \frac{c \times a}{[a b c]} \times \frac{a \times b}{[a b c]} + \frac{a \times b}{[a b c]} \times \frac{b \times c}{[a b c]} \\ = & \frac{[b c a] c - 0}{{[a b c]}^{2}} + \frac{[c a b]}{{[a b c]}^{2}} + \frac{[a b c] b - 0}{{[a b c]}^{2}} \\ = & \frac{a + b + c}{[a + b + c]} \end{array}$