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
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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
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