If ${(a + b + c)}^{2} = 3 (a b + b c + c a)$ , then which one of the following is true?

a \neq b \neq c

No worries! We‘ve got your back. Try BYJU‘S free classes today!

a > b > c

No worries! We‘ve got your back. Try BYJU‘S free classes today!

a < b < c

No worries! We‘ve got your back. Try BYJU‘S free classes today!

a = b = c

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is A $a = b = c$
Given, ${(a + b + c)}^{2} = 3 (a b + b c + c a)$
$\Rightarrow a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a - 3 a b - 3 b c - 3 c a = 0$
$\Rightarrow a^{2} + b^{2} + c^{2} - a b - b c - c a = 0$
$\Rightarrow \frac{1}{2} [{(a - b)}^{2} + {(b - c)}^{2} + {(c - a)}^{2}] = 0$
Which is true only if $a = b = c$

Suggest Corrections

Similar questions

If (a - b), (b - c), (c - a) are in G.P. then prove that $(a + b + c)^{2} = 3 (a b + b c + c a)$

∣ ∣ ∣ ∣ ∣ \begin{matrix} b c - a^{2} & c a - b^{2} & a b - c^{2} - b c + c a + c b & b c - c a + a b & b c + c a - a b (a + b) (a + c) & (b + c) (b + a) & (c + a) (c + b) \end{matrix} ∣ ∣ ∣ ∣ ∣ = 3 (b - c) (c - a) (a - b) (a + b + c) (b c + c a + a b)

a + b + c = 0,

\frac{(b + c)^{2}}{b c} + \frac{(c + a)^{2}}{c a} + \frac{(a + b)^{2}}{a b}

Join BYJU'S Learning Program