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 to a3+b3+c3
Answers
Given : (a+b-5)²+(b+2c+3)²+(c+3a-10)²=0
a,b,c are real numbers
To Find : integer nearest to a³+b³+c³
Solution:
(a+b-5)²+(b+2c+3)²+(c+3a-10)²=0
Square of real number can not be negative
Hence
(a+b-5)² = (b+2c+3)²+ = (c+3a-10)² = 0
(a+b-5)² = 0
=> a + b = 5 Eq1
b + 2c = - 3 Eq2
c + 3a = 10 Eq3
Eq1 - Eq2 => a - 2c = 8
c + 3a = 10 => 2c + 6a = 20
=> 7a = 28
=> a = 4
c = - 2
b = 1
a³+b³+c³ = 4³ + 1³ + (-2)³
= 64 + 1 - 8
= 57
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