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Question

Factorise. (i) 125p3+q3 [2 marks]
(ii) 24a3+81b3 [2 marks]

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Solution

(i) 125p3+q3
=(5p)3+q3
Here, a = 5p and b = q
125p3+q3=(5p+q)[(5p)2(5p)(q)+q2]
...[a3+b3=(a+b)(a2ab+b2)]
(1 mark)
=(5p+q)(25p25pq+q2)
(1 mark)

(ii) 24a3+81b3
[Taking out the common factor 3]
=3[(2a)3+(3b)3]
Here, A =2a and B = 3b
24a3+81b3
=3{(2a+3b)[((2a)2(2a)(3b)+(3b)2]}
...[A3+B3=(A+B)(A2AB+B2)]
(1 mark)
=3(2a+3b)(4a26ab+9b2)
(1 mark)

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