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Question

Multiply each of the following monomials :
(i) 3xyz, 5x, 0 (ii) 65ab,56bc,129abc

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Solution

(i) We have,
(3xyz) × (5x) × 0 = (3×5×0) × (x × x × y × z) = 0 × x2yz = 0

(ii) We have,
(65ab)×(56bc)×(129abc)

=(65×56×129)×(a×a×b×b×b×c×c)=129a(1+1)b(1+1+)c(1+1)=43a2b3c2


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