(i) 6(x + 2y)3 + 8(x + 2y)2
(ii) 14(a – 3b)3 – 21p(a – 3b)
Sol: (i) 6(x + 2y)3 + 8(x + 2y)2
Take out common in all terms,
Then, 2(x + 2y)2 [3(x + 2y) + 4]
Therefore, HCF of 6(x + 2y)3 and 8(x + 2y)2 is 2(x + 2y)2.
(ii) 14(a – 3b)3 – 21p(a – 3b)
14(a – 3b)3 – 21p(a – 3b)
Take out common in all terms,
Then, 7(a – 3b) [2(a – 3b)2 – 3p]
Therefore, HCF of 14(a – 3b)3 and 21p(a – 3b) is 7(a – 3b).
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