(i) 18p2q2 – 24pq2 + 30p2q
(ii) 27a3b3 – 18a2b3 + 75a3b2
Sol: (i) 18p2q2 – 24pq2 + 30p2q
Take out common in all terms,
Then, 6pq (3pq – 4q + 5p)
Therefore, HCF of 18p2q2, 24pq2 and 30p2q is 6pq.
(ii) 27a3b3 – 18a2b3 + 75a3b2
27a3b3 – 18a2b3 + 75a3b2
Take out common in all terms,
Then, 3a2b2 (9ab – 6b + 25a)
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