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If the roots of the equation $x^3-a x^2+b x-c=0$ are in GP, then $\frac{b^3}{a^3}$
Options:
Solution:
2124 Upvotes
Verified Answer
The correct answer is:
$\mathrm{c}$
$$
\begin{aligned}
& \text { Let roots be } \frac{p}{r}, p, p r \\
& \therefore \quad \frac{p}{r}+P+p r=a \\
& \frac{p^2}{r}+P^2 r+p^2=b
\end{aligned}
$$
$$
p^3=c
$$
$$
\Rightarrow \quad \frac{\text { Eq. (ii) }}{\text { Eq. (i) }} \Rightarrow \frac{p^2\left(\frac{1}{r}+r+1\right)}{p\left(\frac{1}{r}+r+1\right)}=\frac{b}{a} \Rightarrow p=\frac{b}{a}
$$
from Eq. (iii) $\left(\frac{b}{a}\right)^3=c$
$$
\frac{b^3}{a^3}=c
$$
Hence, option (3) is correct.
\begin{aligned}
& \text { Let roots be } \frac{p}{r}, p, p r \\
& \therefore \quad \frac{p}{r}+P+p r=a \\
& \frac{p^2}{r}+P^2 r+p^2=b
\end{aligned}
$$
$$
p^3=c
$$
$$
\Rightarrow \quad \frac{\text { Eq. (ii) }}{\text { Eq. (i) }} \Rightarrow \frac{p^2\left(\frac{1}{r}+r+1\right)}{p\left(\frac{1}{r}+r+1\right)}=\frac{b}{a} \Rightarrow p=\frac{b}{a}
$$
from Eq. (iii) $\left(\frac{b}{a}\right)^3=c$
$$
\frac{b^3}{a^3}=c
$$
Hence, option (3) is correct.
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