If 3a + 4b= 9 ,ab = 2 ,find the value of 27a^3 +64b^3
Answers
Answer:
Required numeric value of 27a^3 + 64b^3 is 81.
Step-by-step explanation:
Given, 3a + 4b = 9
Cube on both sides of the equation given above :
= > ( 3a + 4b )^3 = 9^3
From the properties of expansions : -
- ( a + b )^3 = a^3 + b^3 + 3ab( a + b )
Thus,
= > ( 3a )^3 + ( 4b )^3 + 3( 3a x 4b )( 3a + 4b ) = 9^3
= > 27a^3 + 64b^3 + 3( 12 x ab )( 9 ) = 729 { from above, 3a + 4b = 9 }
= > 27a^3 + 64b^3 + 3( 12 x 2 )( 9 ) = 729 { given, ab = 2 }
= > 27a^3 + 64b^3 + 3( 24 )( 9 ) = 729
= > 27a^3 + 64b^3 + 648 = 729
= > 27a^3 + 64b^3 = 729 - 648
= > 27a^3 + 64b^3 = 81
Hence the required numeric value of 27a^3 + 64b^3 is 81.
Answer:
81
Step-by-step explanation:
3a + 4b = 9
Cubing both the sides;
( 3a + 4b )³ = 9³
⇒ (3a)³ + (4b)³ + 3 * 3a *4b [ 3a + 4b ] = 729
⇒ 27a³ + 64b³ + 3* 12 * 2 (9) = 729
[ ∵ ab = 2 and 3a + 4b = 9]
⇒ 27a³ + 64b³ + 648 = 729
⇒ 27a³ + 64b³ = 729 - 648
⇒ 27a³ + 64b³ = 81
Hence, the answer is 81.
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