(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)(a2−ab+b2)]
(1 mark)
=(5p+q)(25p2−5pq+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)(A2−AB+B2)]
(1 mark)
=3(2a+3b)(4a2−6ab+9b2)
(1 mark)