If roots of the equation $(c^2 - a b)x^2 - 2(a^2 - b c)x + (b^2 - a c) = 0$ are equal, prove that either $a = 0$ or $a^3 + b^3 + c^3 = 3abc$.
Step-by-Step Solution
Key Concept: $D = 4(a^2 - bc)^2 - 4(c^2 - ab)(b^2 - ac) = 0 \Rightarrow (a^4 - 2a^2 bc + b^2 c^2) - (b^2 c^2 - ac^3 - ab^3 + a^2 bc) = 0 \Rightarrow a^4 - 3a^2 bc + ac^3 + ab^3 = 0 \Rightarrow a(a^3 + b^3 + c^3 - 3abc) = 0 \Rightarrow a = 0$ or $a^3 + b^3 + c^3 = 3abc$.
$D = 4[(a^2 - bc)^2 - (c^2 - ab)(b^2 - ac)] = 0$. [1.0 Mark]
Expand: $a^4 - 2a^2 bc + b^2 c^2 - (b^2 c^2 - ac^3 - ab^3 + a^2 bc) = 0 \Rightarrow a^4 + ab^3 + ac^3 - 3a^2 bc = 0$. [1.0 Mark]
Factor out $a$: $a(a^3 + b^3 + c^3 - 3abc) = 0 \Rightarrow a = 0$ or $a^3 + b^3 + c^3 = 3abc$. Proved! [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Setting up $D = 0$: 1.0 Mark
Expanding and simplifying: 1.0 Mark
Factoring out $a$ to get $a=0$ or $a^3+b^3+c^3=3abc$: 1.0 Mark
Correct Answer: