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

Question

Find the square root of $4a^2 + 9b^2 + c^2 + 12ab - 6bc - 4ac$.

Answer 1

We use the formula: Cube of a sum: $ (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 $ Cube of a difference: $ (a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 $

Step 1: Express 99 as $ 100 - 1 $ \[ (99)^3 = (100 - 1)^3 \]

Step 2: Use $ (a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 $, where $ a=100, b=1 $

$ 100^3 - 3(100^2)(1) + 3(100)(1^2) - 1^3 $

$ 1000000 - 30000 + 300 - 1 = 970299 $

Step 1: Express 102 as $ 100 + 2 $ \[ (102)^3 = (100 + 2)^3 \]

Step 2: Use $ (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 $, where $ a=100, b=2 $

$ 100^3 + 3(100^2)(2) + 3(100)(2^2) + 2^3 $

$ 1000000 + 60000 + 1200 + 8 = 1061208 $

Step 1: Express 998 as $ 1000 - 2 $ \[ (998)^3 = (1000 - 2)^3 \]

Step 2: Use $ (a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 $, where $ a=1000, b=2 $

$ 1000^3 - 3(1000^2)(2) + 3(1000)(2^2) - 2^3 $

$ 1000000000 - 6000000 + 12000 - 8 = 994012992 $

Solution and Explanation