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

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Similar Questions

Let $S$ be the set of all $\alpha \in R$ such that the equation, $cos\,2 x + \alpha \,sin\, x = 2\alpha -7$ has a solution. Then $S$ is equal to

The number of distinct real roots of the equation $x^{5}\left(x^{3}-x^{2}-x+1\right)+x\left(3 x^{3}-4 x^{2}-2 x+4\right)-1=0$ is

The number of solution$(s)$ of the equation $ln(lnx)$ = $log_xe$ is -

If $x$ is real, then the value of ${x^2} - 6x + 13$ will not be less than