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Question
Verify the following:
(ab + bc)(ab – bc) + (bc + ca)(bc – ca) + (ca + ab)(ca – ab) = 0
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Solution
Taking L.H.S. = (ab + bc)(ab – bc) + (bc + ca)(bc – ca) + (ca + ab)(ca – ab)
= [(ab)2 – (bc)2] + [(bc)2 – (ca)2] + [(ca)2 – (ab)2] ...[Using the identity, (a + b)(a – b) = a2 – b2]
= a2b2 – b2c2 + b2c2 – c2a2 + c2a2 – a2b2
= 0
= R.H.S. ...[Cancelling the like terms having opposite signs]
Hence verified.
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