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

b = 0

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a³ + b³ + c³ = 3abc

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a³ + b³ + c³ = abc

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a = 0

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Solution

The correct options are
A
b = 0

B
a³ + b³ + c³ = 3abc

Roots of the equation are equal
$\Rightarrow$ Discriminant = 0
$D = (2 (b^{2} - a c))^{2} - 4 (a^{2} - b c) (c^{2} - a b)$
$= 4 [(b^{2} - a c)^{2} - (a^{2} - b c) (c^{2} - a b)]$
$= 4 [b^{4} + a^{2} c^{2} - 2 b^{2} a c - a^{2} c^{2} + a^{3} b - b^{2} a c + b c^{3}]$
$= 4 [b^{4} + a^{3} b + b c^{3} - 3 b^{2} a c]$
D=0
$\Rightarrow b (b^{3} + a^{3} + c^{3} - 3 a b c) = 0$
$\Rightarrow b = 0 o r b^{3} + a^{3} + c^{3} - 3 a b c = 0$

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