If a,b,c are distinct real numbers and | vedclass

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Question

$ a^{3}+b^{3}+c^{3}-3 a b c =0 $

$\frac{1}{2}(a+b+c)\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right] =0 $

$ \Rightarrow a+b+c=0$

$\because a, b, c$

Vedclass · Accepted Answer

$\frac {c}{a}$

Vedclass · Answer

$ a^{3}+b^{3}+c^{3}-3 a b c =0 $

$\frac{1}{2}(a+b+c)\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right] =0 $

$ \Rightarrow a+b+c=0$

$\because a, b, c$ are distinct

Now equation $a x^{2}+b x+c=0$ has one root $1$.

Now, other root is $\frac{\mathrm{c}}{\mathrm{a}}$

If $a,b,c$ are distinct real numbers and $a^3 + b^3 + c^3 = 3abc$ , then the equation $ax^2 + bx + c = 0$ has two roots, out of which one root is

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