prove that [a*b b*c c*a]=[abc]^2?

Answer 1

a*b*b*c*c*a
=a*b*c*a*b*c as multiplication is commutative and associative
=(abc)(abc)
=(abc)^2

Answer 2

[a*b b*c c*a]= [a * b * b * c * c * a]
= a^2 * b^2 * c^2
= [a * b * c] * [a * b * c]
= [abc] ^2

Thus proved

Answer 3

Given, [a*b b*c c*a]=[abc]^2
Now, divide the given equation by [abc]^2
Then we have, [a*b b*c c*a]/[abc]^2 = 1
Also,[a*b b*c c*a]/[abc] [bca] = 1 ; and on multiplication of the denominator we get, [a*b b*c c*a]/[a*b b*c c*a] =1
Consequently, 1=1
Thus proved.

Answer 4

[a*b][ b*c][ c*a]
=[a*a][b*b][c*c]
=a^2*b^2*c^2
=[abc]^2

Answer 5

[a*b b*c c*a]
=[ab bc ca]
=[abc bc a]
=[abc abc]
=[abc][abc]
=[abc]^2

Answer 6

[a*b * b*c * c*a]= [ab * bc * ca]
[aa*bb*cc]=[abc]*[abc]
[abc]^2