Question

If a, b, c are real numbers and (a+b-5) ² + (b+2c+3) ² + (c+3a-10) ² =0
Find the integer nearest a³+b³+c³​

Answer 1

Given :  a, b, c are real number and (a+b-5) ²+(b+2c+3) ²+(c+3a-10) ² = 0

To Find : the integer nearest to a³+b³+c³​

Solution:

(a+b-5) ²+(b+2c+3) ²+(c+3a-10) ² = 0

Sum of squares = 0 hence each expression in square is 0

a+b-5 = 0

=> a + b = 5

b+2c+3 = 0

=> b + 2c   = 3

c+3a-10  = 0

=> c  +  3a = 10  

=> c  = 10 - 3a

b + 2c   = 3

=> b + 2(10 - 3a ) = 3

=> b + 20 - 6a = 3

=> 6a - b = 17

a + b = 5

6a - b = 17

=> 7a  = 22

=> a = 22/7

 b = 13/7

 c   = 4/7

a³+b³+c³​  = (22/7)³  + (13/7)³  + (4/7)³

= ( 22³  + 12³  + 4³)/7³

= ( 10648 + 1728 + 64)/343

= 12440/343

= 36.27

= 36

integer nearest to a³+b³+c³​ = 36