**(i) 10a(2p + q)**^{3} – 15b (2p + q)^{2} + 35 (2p + q),

^{3}– 15b (2p + q)

^{2}+ 35 (2p + q),

**(ii) x(x**^{2} + y^{2} – z^{2}) + y(-x^{2} – y^{2} + z^{2}) – z (x^{2} + y – z^{2})

^{2}+ y

^{2}– z

^{2}) + y(-x

^{2}– y

^{2}+ z

^{2}) – z (x

^{2}+ y – z

^{2})

**Sol: **(i) 10a(2p + q)^{3} – 15b (2p + q)^{2} + 35 (2p + q)

Take out common in all terms,

Then, 5(2p + q) [2a (2p + q)^{2} – 3b (2p + q) + 7]

Therefore, HCF of 10a(2p + q)^{3}, 15b (2p + q)^{2} and 35 (2p + q) is 5(2p + q).

**(ii) **x(x^{2} + y^{2} – z^{2}) + y(-x^{2} – y^{2} + z^{2}) – z (x^{2} + y – z^{2})

x(x^{2} + y^{2} – z^{2}) + y(-x^{2} – y^{2} + z^{2}) – z (x^{2} + y – z^{2})

Take out common in all terms,

Then, (x^{2} + y^{2} – z^{2}) [x – y – z]

Therefore, HCF of x(x^{2} + y^{2} – z^{2}), y(-x^{2} – y^{2} + z^{2}) and z (x^{2} + y – z^{2}) is (x^{2} + y^{2} – z^{2})

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