Question

027. Evaluates (102) using suitable identity​

Answer 1

Answer:

=(100)3(3)3+3× 100×3×(100+3) [using identity ,(a+b)3=a3+b3+ 3ab(a+b)] =10000000+27+900(103

Answer 2

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( 102 )³ can be written as ( 100 + 2 )³

Using identity here ;

( a + b )³ = a³ + b³ + 3ab ( a + b )

⇒ ( 100 + 2 )³

⇒ (100)³ + (2)³ + 3 * 100 * 2 ( 100 + 2 )

⇒ 1000000 + 8 + 600 ( 102 )

⇒ 1000000 + 8 + 61200

⇒ 1061208

Therefore the answer is 1061208.