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If $a.b $ & $c$ are rational number satisfying the equation $x^3+ax^2+bx+c=0$, then which of the following can also be true .

A) $a+b^2+c^3=0$

B) $a+b^2+c^3=5$

C) $a+b^2+c^3=1$

D) None of these

If it is a cubic function the equation has at least one real root. So for any value of a, b & c we have one real root, hence i am confused. The question i presume is cotrect

Samar Imam Zaidi
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    If the equation is supposed to hold for all real numbers $x$ then the answer is A). If it holds for just one $x$ then thr answer is D). – Kavi Rama Murthy Aug 12 '21 at 08:23
  • Vieta's formulas might be interesting to you. As @KaviRamaMurthy says, there's not enough context here, and there exist exactly $3$ real or complex $x$ for which the equation turns out to be zero, regardless of what $a,b,c$ are – FShrike Aug 12 '21 at 08:24

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Since no relation between $a, b$ and $c$ need to hold in order to satisfy $$x^3+ax^2+bx+c=0 \tag{i}$$ as there always exist some real $x$ which is one of the roots of $\text{(i)}$, I'm assuming that $a, b$ and $c$ satisfy $\text{(i)}$ for all $x \in \mathbb R$

Thus, for $x=0$, equation $\text{(i)}$ becomes

$0+0+0+c=0 \implies c=0$

Also, $1+a+b+c=0$ and $ -1+a-b+c=0$ for $x=1$ and $x=-1$ in $\text{(i)}$ respectively.

Solving them will give you $a=0,b=-1,c=0$

But for $x=2$,we get $8+4a+2b+c=0 \implies 8-2=0$ which is a contradiction, so there do not exists any rational numbers $a,b,c$ that can satisfy $\text{(i)}$ for all $x \in \mathbb R$

Aman Kushwaha
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