Proof using Permutation Symbols

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BlazNProdigy
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Homework Statement


Proove that...
(AxB)x(CxD)=(A.BxD)C-(A.BxC)D=(A.CxD)B-(B.CxD)A
using Permutation Symbols

Homework Equations

The Attempt at a Solution


I am confused about what to do after the third line from 'vela's response' (Post #2 from the reference link below).

Reference https://www.physicsforums.com/threa...rmutation-tensor-and-kroenecker-delta.454568/
 
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The problem is straight-forward, just tedious. You just apply the following rules:

  1. [itex](X \times Y)^c = \epsilon_{abc} X^a Y^b[/itex]
  2. [itex](X \cdot Y) = \delta_{ab} X^a Y^b[/itex]
  3. [itex]\epsilon_{abc} = \epsilon_{bca} = \epsilon_{cab} = -\epsilon_{bac} = -\epsilon{acb} = - \epsilon_{cba}[/itex]
  4. [itex]\epsilon_{abc} \epsilon_{ade} = \delta_{bd} \delta_{ce} - \delta_{be} \delta_{cd}[/itex]
  5. [itex]\epsilon_{abc} \delta_{ae} = \epsilon_{ebc}[/itex]
To get the ball rolling, rewrite what you're being asked to prove in terms of components:

[itex]((A \times B) \times (C \times D))^c = (A \cdot (B \times D)) C^c - (A \cdot (B \times C)) D^c = (A \cdot (C \times D)) B^c - (B \cdot (C \times D)) A^c[/itex]
 
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