Question:
prove that [a*b , b*c , c*a] = [abc]^2
Answer:
[a * b, b * c, c * a] = {(a * b) * (b * c)} . (c * a)
=> [a * b, b * c, c * a] = {d * (b * c)} . (c * a) {here d = a * b}
=> [a * b, b * c, c * a] = {(d . c)b - (d * b)c} . (c * a)
=> [a * b, b * c, c * a] = {((a * b) . c)b - ((a * b) * b)c} . (c * a)
=> [a * b, b * c, c * a] = {[a b c]b - [a b b]c} . (c * a)
=> [a * b, b * c, c * a] = {[a b c]b - 0*c} . (c * a) {since [a b b] = 0}
=> [a * b, b * c, c * a] = {[a b c]b} . (c * a)
=> [a * b, b * c, c * a] = [a b c]{b . (c * a)}
=> [a * b, b * c, c * a] = [a b c][b c a]} {since b . (c * a) = [b c a]}
=> [a * b, b * c, c * a] = [a b c][a b c] {since [b c a] = [a b c]}
=> [a * b, b * c, c * a] = [a b c]2
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