Evaluate the following using suitable identities:
(i) $ (99)^{3} $
(ii) $ (102)^{3} $
(iii) $ (998)^{3} $

AcademicMathematicsNCERTClass 9

To do:

We have to evaluate the given expressions using suitable identities.

Solution:

We know that,

$(a+b)^3=a^3 + b^3 + 3ab(a+b)$

$(a-b)^3= a^3-b^3-3ab(a-b)$

Therefore,

(i) $(99)^3 = (100 - 1)^3$

$= (100)^3 - (1)^3 - 3 \times 100 \times 1 (100 - 1)$

$= 1000000 - 1 - 300 \times 99$

$= 1000000 - 1 - 29700$

$= 1000000 - 29701$

$= 970299$

Hence, $(99)^3 = 970299$.   

 (ii) $(102)^3 = (100 + 2)^3$

$= (100)^3 + (2)^3 + 3 \times 100 \times 2 (100 + 2)$

$= 1000000 + 8 + 600 \times (100+2)$

$= 1000008 + 600\times100 + 600\times2$

$= 1000008 + 60000 + 1200$

$=1061208$

Hence, $(102)^3 = 1061208$.

(iii)   $(998)^3 = (1000 - 2)^3$

$= (1000)^3 - (2)^3 - 3 \times 1000 \times 2 (1000 - 2)$

$= 1000000000 - 8 - 6000 \times (1000-2)$

$= 999999992 - 6000000+12000$

$= 999999992 -5988000$

$= 994011992$

Hence, $(998)^3 = 994011992$.   

raja
Updated on 10-Oct-2022 13:39:07

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