#### (i) 15a (2p – 3q) – 10b (2p – 3q)

#### (ii) 3a(x^{2} + y^{2}) + 6b (x^{2} + y^{2})

**Sol: **(i) 15a (2p – 3q) – 10b (2p – 3q)

Take out common in both terms,

Then, 5(2p – 3q) [3a – 2b]

Therefore, HCF of 15a (2p – 3q) and 10b (2p – 3q) is 5(2p – 3q).

##### (ii) 3a(x^{2} + y^{2}) + 6b (x^{2} + y^{2})

3a(x^{2} + y^{2}) + 6b (x^{2} + y^{2})

Take out common in all terms,

Then, 3(x^{2} + y^{2}) (a + 2b)

Therefore, HCF of 3a(x^{2} + y^{2}) and 6b (x^{2} + y^{2}) is 3(x^{2} + y^{2}).

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