Question

a, b, c ∈ R and $a^2+b^2+c^2=ab+bc+ca$ then,

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:a, \: b, \: c \: \in \: R$

and

$\rm :\longmapsto\: {a}^{2} + {b}^{2} + {c}^{2} = ab + bc + ca$

can be rewritten as

$\rm :\longmapsto\: {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca = 0$

On multiply by 2 on both sides, we get

$\rm :\longmapsto\: 2{a}^{2} +2{b}^{2} + 2{c}^{2} - 2ab - 2bc - 2ca = 0$

$\rm :\longmapsto\: {a}^{2} + {a}^{2} +{b}^{2} + {b}^{2} + {c}^{2} + {c}^{2} - 2ab - 2bc - 2ca = 0$

On rearranging the terms, we get

$\rm :\longmapsto\:( {a}^{2} + {b}^{2} - 2ab) + ( {b}^{2} + {c}^{2} - 2bc) + ( {c}^{2} + {a}^{2} - 2ac) = 0$

$\rm :\longmapsto\: {(a - b)}^{2} + {(b - c)}^{2} + {(c - a)}^{2} = 0$

$\: \: \: \: \: \: \: \: \red{\bigg \{ \because \: {x}^{2} + {y}^{2} - 2xy = {(x - y)}^{2} \bigg \}}$

We know, sum of squares is 0 only, when number itself is 0.

$\rm :\implies\:a - b = 0 \: \: and \: \: b - c =0 \: \: and \: c - a = 0$

$\bf\implies \:a = b = c$

Hence,

$\rm :\longmapsto\: If \: {a}^{2} + {b}^{2} + {c}^{2} = ab + bc + ca \: then \: a = b = c$

More Identities to know:

• (a + b)² = a² + 2ab + b²

• (a - b)² = a² - 2ab + b²

• a² - b² = (a + b)(a - b)

• (a + b)² = (a - b)² + 4ab

• (a - b)² = (a + b)² - 4ab

• (a + b)² + (a - b)² = 2(a² + b²)

• (a + b)³ = a³ + b³ + 3ab(a + b)

• (a - b)³ = a³ - b³ - 3ab(a - b)