POLYNOMIALS 47
So, we get the following identity:
Identity VI : (x + y)^3 =x^3 + y^3 + 3xy (x + y)
Also, by replacing y by –y in the Identity VI, we get
Identity VII : (x – y)^3 =x^3 – y^3 – 3xy(x – y)
=x^3 – 3x^2 y + 3xy^2 – y^3Example 22 : Write the following cubes in the expanded form:
(i) (3a + 4b)^3 (ii) (5p – 3q)^3Solution : (i) Comparing the given expression with (x + y)^3 , we find that
x =3a and y = 4b.So, using Identity VI, we have:
(3a + 4b)^3 =(3a)^3 + (4b)^3 + 3(3a)(4b)(3a + 4b)
=27a^3 + 64b^3 + 108a^2 b + 144ab^2(ii) Comparing the given expression with (x – y)^3 , we find that
x =5p, y = 3q.So, using Identity VII, we have:
(5p – 3q)^3 =(5p)^3 – (3q)^3 – 3(5p)(3q)(5p – 3q)
= 125p^3 – 27q^3 – 225p^2 q + 135pq^2Example 23 : Evaluate each of the following using suitable identities:
(i) (104)^3 (ii)(999)^3Solution : (i) We have
(104)^3 = (100 + 4)^3
= (100)^3 + (4)^3 + 3(100)(4)(100 + 4)
(Using Identity VI)
= 1000000 + 64 + 124800
= 1124864(ii) We have
(999)^3 = (1000 – 1)^3
= (1000)^3 – (1)^3 – 3(1000)(1)(1000 – 1)
(Using Identity VII)
= 1000000000 – 1 – 2997000
= 997002999