Question

if 4a2+9b2+16c2+(1/9a2)+(1/16b2)+(1/25c2)=133/30 then find 1440a2b2c2

Answer 1

Answer:

1440a²b²c² = 1

Step-by-step explanation:

4a² + 9b² + 16c²  + 1/9a² + 1/16b²  + 1/25c²  = 133/30

=> 4a² + 1/9a² + 9b² + 1/16b² + 16c² + 1/25c²  = 133/30

=> (2a)² + (1/3a)²  + (3b)² + (1/4b)²  + (4c)²  + (1/5c)² = 133/30

=> (2a - 1/3a)² + 4/3  + (3b - 1/4b)² + 3/2  + (4c - 1/5c)² + 8/5 = 133/30

=>  (2a - 1/3a)²  + (3b - 1/4b)² + (4c - 1/5c)²  + 133/30  = 133/30

=> (2a - 1/3a)²  + (3b - 1/4b)² + (4c - 1/5c)²  = 0

2a - 1/3a = 0  , 3b - 1/4b = 0  , 4c - 1/5c = 0

=> 2a = 1/3a  => 6a² = 1

3b = 1/4b => 12b² = 1

4c = 1/5c => 20c² = 1

6a² * 12b² * 20c² = 1 * 1 * 1

=> 1440a²b²c² = 1