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Question

If $$c^{2}\neq ab$$, and the root of$$ (c^{2}-ab)x^{2}-2(a^{2}-bc)x+(b^{2}-ac)=0$$ are equal then show that $$a^{3}+b^{3}+c^{3}=3abc $$ or a =0


Solution

$$(c^{2}-ab) x^{2}-2 (a^{2}-bc) x+ (b^{2}-ac)=0$$
It is of form $$Ax^{2}+Bx+ C=0$$
For the roots to be equal 
$$\Rightarrow B^{2}= 4AC$$
$$4 (a^{2}-bc)^{2}= 4 (c^{2}-ab) (b^{2}- ac)$$
$$\Rightarrow a^{5}+ b^{2}c^{2}- 2a^{2}bc = b^{2} c^{2}- ac^{3}-ab^{3}+a^{2} bc$$
$$\Rightarrow a^{4}+ ac^{3}+ab^{3}= 3 a^{2} bc$$
$$\Rightarrow a (a^{3}+ b^{3} + c^{3})= a (3abc)$$
$$\Rightarrow \ Either\ a=0 $$ or if $$a \neq 0, a^{3}+b^{3}+ c^{3}= 3abc$$

Mathematics

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