Question

If a^2 + b^2 + c^2 = ab + bc + ca, find the value of
(а + с)/b=?​

Answer 1

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

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

$\large\underline{\sf{To\:Find - }}$

$\rm :\longmapsto\:\dfrac{a + c}{b}$

Solution :-

Given that,

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

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

☆ On multiply by 2, both sides, we get

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

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

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

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

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

Now,

We know that, Sum of squares is zero iff term itself is 0.

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

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

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

Now,

Consider,

$\rm :\longmapsto\:\dfrac{a + c}{b}$

$\rm \: = \: \: \dfrac{a + a}{a}$

$\red{\bigg \{ \because \: \sf \: a = b = c\bigg \}}$

$\rm \: = \: \: \dfrac{2a}{a}$

$\rm \: = \: \: 2$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \underbrace{\boxed{\bf\implies \:\:\dfrac{a + c}{b} = 2}}$

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)