If $a = x + y, b = x - y, c = x + 2 y,$ then the value of $a^{2} + b^{2} + c^{2} - a b - b c - c a$ is

4 y^{2}

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

5 y^{2}

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

6 y^{2}

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

7 y^{2}

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

Open in App

Solution

The correct option is D $7 y^{2}$
$a - b = x + y - x + y = 2 y$
$b - c = x - y - x - 2 y = - 3 y$
$c - a = x + 2 y - x - y = y$
Now, $a^{2} + b^{2} + c^{2} - a b - b c - c a$
= $\frac{1}{2} (a - b)^{2} + (b - c)^{2} + (c - a)^{2}$
= $\frac{1}{2} (2 y)^{2} + (- 3 y)^{2} + y^{2} = \frac{1}{2} \times 14 y^{2} = 7 y^{2}$

Suggest Corrections

Similar questions

x = a^{2} - b c, y = b^{2} - c a, z = c^{2} + a b

\frac{(a + b + c) (x + y + z)}{a x + b y + c z}

x = b + c, y = c + a, z = a + b

\frac{x^{2} + y^{2} + z^{2} - y z - z x - x y}{a^{2} + b^{2} + c^{2} - b c - c a - a b}

a^{2} + b^{2} + c^{2} - a b - a c - b c

a = x + y, b = x - y

c = x + 2 y

If $a + b + c = 0$ then find value of $\frac{a^{2}}{b c} + \frac{b^{2}}{a c} + \frac{c^{2}}{a b}$

Join BYJU'S Learning Program