If $a^{3} + b^{3} + c^{3} - 3 a b c = 0$ then the roots of the equation
$(a^{2} - b c) x^{2} + 2 (b^{2} - a c) x + c^{2} - a b = 0$ are

imaginary

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

real and unequal

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

real and equal

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

Cannot say

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

Open in App

Solution

The correct option is D real and equal
Given quadratic equation is $(a^{2} - b c) x^{2} + 2 (b^{2} - a c) x + (c^{2} - a b) = 0$ and $a^{3} + b^{3} + c^{3} = 3 a b c$
$D = 4 {(b^{2} - a c)}^{2} - 4 (a^{2} - b c) (c^{2} - a b) \Rightarrow D = 4 b (a^{3} + b^{3} + c^{3} - 3 a b c) = 0$
$∴$ Roots are real and equal.

Suggest Corrections

Similar questions

If the roots of the equation $(a^{2} - b c) x^{2} + 2 (b^{2} - a c) x + c^{2} - a b = 0$ are equal, then

c^{2} \neq a b

(c^{2} - a b) x^{2} - 2 (a^{2} - b c) x + (b^{2} - a c) = 0

a^{3} + b^{3} + c^{3} = 3 a b c

(c^{2} - a b) x^{2} - 2 (a^{2} - b c) x + b^{2} - a c = 0

a^{3} + b^{3} + c^{3} = 3 a b c

(a_{1}, a_{2}, a_{3})

(b_{1}, b_{2}, b_{3})

(c_{1}, c_{2}, c_{3})

x (a_{1}, a_{2}, a_{3}) + y (b_{1}, b_{2}, b_{3}) + z (c_{1}, c_{2}, c_{3}) = 0

\frac{a^{3} + b^{3} + c^{3} - 3 a b c}{a^{2} + b^{2} + c^{2} - a b - b c - c a}

Join BYJU'S Learning Program