Solveeit Logo
Login

Question

Question: If the equation \({a^2} + {b^2} + {c^2} - ab - bc - ac = 0\) then prove that \(a = b = c\)....

If the equation a2+b2+c2−ab−bc−ac=0{a^2} + {b^2} + {c^2} - ab - bc - ac = 0 then prove that a=b=ca = b = c.

Explanation

Solution

Hint: In order to solve this question multiply 2 with the equation and try to make the formula of (a−b)2, (b−c)2, (a−c)2{(a - b)^2},\,{(b - c)^2},\,{(a - c)^2} by solving the given equation, then you will get solution to this problem.

Complete step-by-step answer:

The given equation is a2+b2+c2−ab−bc−ac=0{a^2} + {b^2} + {c^2} - ab - bc - ac = 0 ……(1)
We have to prove: a=b=ca = b = c
On multiplying 2 with the given equation we get the equation as:
2(a2+b2+c2−ab−bc−ac)=2(0) 2a2+2b2+2c2−2ab−2bc−2ac=0  2({a^2} + {b^2} + {c^2} - ab - bc - ac) = 2(0) \\\ 2{a^2} + 2{b^2} + 2{c^2} - 2ab - 2bc - 2ac = 0 \\\
On solving it further we get the equation as:
a2+a2+b2+b2+c2+c2−2ab−2bc−2ac=0 a2−2ab+b2+a2−2ac+c2+b2−2bc+c2=0  {a^2} + {a^2} + {b^2} + {b^2} + {c^2} + {c^2} - 2ab - 2bc - 2ac = 0 \\\ {a^2} - 2ab + {b^2} + {a^2} - 2ac + {c^2} + {b^2} - 2bc + {c^2} = 0 \\\
We know the formulas as :
(a−b)2=a2−2ab+b2, (b−c)2=b2−2bc+c2, (a−c)2= a2−2ac+c2{(a - b)^2} = {a^2} - 2ab + {b^2},\,{(b - c)^2} = {b^2} - 2bc + {c^2},\,{(a - c)^2} = \,{a^2} - 2ac + {c^2} …(2)
Putting the values of equation (2) in obtained equation we get the new equation as:
(a−b)2+ (b−c)2+(a−c)2=0{(a - b)^2} + \,{(b - c)^2} + {(a - c)^2} = 0
Since the sum of square is zero then each term should be zero
(a−b)2=0{(a-b)^2=0} , (b−c)2=0{(b-c)^2=0} , (c−a)2=0{(c-a)^2=0}
(a−b)=0{(a-b)=0}, (b−c)=0{(b-c)=0}, (c−a)=0{(c-a)=0}
a=b{a=b}, b=c{b=c}, c=a{c=a}
The above shown condition is only possible when a=b=ca = b = c.
Hence, proved.

Note: Whenever you are struck with these types of problems you should always think about which identity we can use so that we can prove what has been asked like here we have to make the given equation such that we can use the formula of (a−b)2{(a - b)^2}. Proceeding like this will make your solution correct and proved.