Question

If a2 +b2 +c2 -ab - bc -ca =0 then show that a=b=c​

Answer 1

$\huge{\orange{\underline{\underline{\bf{AnSweR:}}}}}$

Given,

a²+b²+c²-ab-bc-ca = 0

Multiple With "2" on both sides

➝ $\sf{2(a²+b²+c²-ab-bc-ca) = 2(0)}$

➝$\sf {2a²+2b²+2c²-2ab-2bc-2ca = 0}$

➝ $\sf {(a²-2ab+b²)+(b²-2bc+c²)+(c²-2ca+a²) = 0}$

$\boxed{\bf{(x-y)² = x²-2xy+y²}}$

➝ $\sf {(a-b)²+(b-c)²+(c-a)² = 0}$

As, The Sum of the Squares is Zero '0' then each term can be equated with '0'.

➝ $\sf {(a-b)² = 0, (b-c)² = 0, (c-a) = 0}$

➝ $\sf {(a-b) = 0, (b-c) = 0, (c-a) = 0}$

➝ $\sf {a=b, b=c, c=a}$

☞ $\red{\sf{a = b = c}}$

Hence, Proved.

Hope It Helps You ✌️

Answer 2

Given,

a²+b²+c²-ab-bc-ca = 0

Multiple With "2" on both sides

➝ $\sf{2(a²+b²+c²-ab-bc-ca) = 2(0)}$

➝$\sf {2a²+2b²+2c²-2ab-2bc-2ca = 0}$

➝ $\sf {(a²-2ab+b²)+(b²-2bc+c²)+(c²-2ca+a²) = 0}$

$\boxed{\bf{(x-y)² = x²-2xy+y²}}$

➝ $\sf {(a-b)²+(b-c)²+(c-a)² = 0}$

As, The Sum of the Squares is Zero '0' then each term can be equated with '0'.

➝ $\sf {(a-b)² = 0, (b-c)² = 0, (c-a) = 0}$

➝ $\sf {(a-b) = 0, (b-c) = 0, (c-a) = 0}$

➝$\sf {a=b, b=c, c=a}$

$\gray{\sf{a = b = c}}$

.$\rm\underline\bold{Hence, Proved \red{\huge{\checkmark}}}$