Polynomials

Exercise 2.5 Part 8

Question 7: Evaluate the following using suitable identities:

(i) `99^3`

Answer: Given, `99^3=(100 – 1)^3`

Let, `x = 100` and `y = 1`

Using the identity `(x – y)^3= x^3 - y^3 - 3x^2y + 3xy^2`

We get;
`100^3 - 1^3 - 3 xx (100^2) xx 1 + 3 xx 100 xx 1^2`

`= 1000000 – 1 -30000 + 300`

`= 1000000 + 300 – 1 – 30000 = 970299`


(ii) `102^3`

Answer: Given; `102^3= (100 + 2)^3`

Let, `a = 100` and `b = 2`

Using the identity `(a + b)^3= a^3+ b^3 + 3ab(a + b)`

We get; `100^3 + 2^3 + 3 xx 100 xx 2(100 + 2)`

`= 1000000 + 8 + 600 xx 102`

`= 1000000 + 8 + 61200= 1000008 + 61200 = 1061208`


(iii) `998^3`

Answer: Given; `998^3= (1000 – 2)^3`

Let, `a = 1000` and `b = 2`

Using the identity `(a – b)^3= a^3 - b^3 - 3a^2\b + 3ab^2`

We get; `1000^3 - 2^3 - 3 xx 1000^2\ xx 2 + 3 xx 1000 xx 2^2`

`= 1000000000 – 8 – 6000000 + 12000= 994011992`



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