Question

117. For real a, b, c, if a² + b²+ c² = ab + bc + ca,
the value of is
a+c/b
​

Answer 1

Answer:

add 2ab+2bc+2ac on both sides

we get,

(a+b+c) ²=3ab+3bc+3ca

{(a+b+c) ²-3ca}/3b=a+c

divide by b on both sides

{(a+b+c) ²-3ca}/3b*1/b=a+c/b

{(a+b+c) ²-3ca}/3b²=a+c/b

Answer 2

Answer:

Given that,

• a² + b² + c² = ab + bc + ac

On first thought, you would come across the identities

• (a+b)²
• (a-b)²

But since, we get '-ab' while transposing to the RHS, we will try to convert everything to the form of (a-b)² type.

Let' first multiply by 2 on both sides. Hence we get:

→ 2 ( a² + b² + c² ) = 2 ( ab + bc + ac )

→ 2a² + 2b² + 2c² = 2ab + 2bc + 2ac

Transposing the terms from RHS to the LHS we get:

→ 2a² + 2b² + 2c² - 2ab - 2bc - 2ac = 0

Splitting 2a² as 'a²' and 'a²', we get:

→ a² + a² + b² + b² + c² + c² - 2ab - 2bc - 2ac = 0

Now rearranging terms we get:

→ ( a² - 2ab + b² ) + ( b² - 2bc + c² ) + ( c² - 2ac + a² ) = 0

→ ( a - b )² + ( b - c )² + ( c - a )² = 0

Now the sum of squares of 3 numbers will only be zero, if all the three numbers are individually zero. Hence we get,

• ( a - b ) = 0
• ( b - c ) = 0
• ( c - a ) = 0

From the above we can imply that,

⇒ a = b ; b = c ; c = a

Hence a = b = c is the required condition.

Now we are required to find the value of:

$\boxed{ \dfrac{a+c}{b} = ?}$

Hence taking 'b' and 'c' equal to 'a' we get:

$\implies \dfrac{a+a}{a}\\\\\\\implies \dfrac{2a}{a} = \boxed{2}$

Hence the required value is 2.