Question

If a > b > c and a, b, c ∈ R such that a3 + b3 + c3 = 3abc, then the equation ax2 + bx + c = 0 has

Answer 1

$\Hugex=1$

$\underline{\underline{\text{Explanation}}}$

Given that,

$\cdots\longrightarrow a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)=0.$

As we know that three real values are different,

$\cdots\longrightarrow a^{2}+b^{2}+c^{2}-ab-bc-ca=\dfrac{1}{2}\{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\}\neq0$

the equation has $\underline{a+b+c=0}$ as a solution.

Let us substitute $x=1$.

$\cdots\longrightarrow\underline{a+b+c=0.}$

The equation has $\boxed{x=1}$ as a solution.