Q115 / 160 · AP EAMCET 2020 (22 Sep Shift 2) · Mathematics

If the roots of the equation $x^3-a x^2+b x-c=0$ are in GP, then $\frac{b^3}{a^3}$

  • A
    1
  • B
    –1
  • Correct answer
    C
    $\mathrm{c}$
  • D
    $-c$

Solution

$$ \begin{aligned} & \text { Let roots be } \frac{p}{r}, p, p r \\ & \therefore \quad \frac{p}{r}+P+p r=a \\ & \frac{p^2}{r}+P^2 r+p^2=b \end{aligned} $$ $$ p^3=c $$ $$ \Rightarrow \quad \frac{\text { Eq. (ii) }}{\text { Eq. (i) }} \Rightarrow \frac{p^2\left(\frac{1}{r}+r+1\right)}{p\left(\frac{1}{r}+r+1\right)}=\frac{b}{a} \Rightarrow p=\frac{b}{a} $$ from Eq. (iii) $\left(\frac{b}{a}\right)^3=c$ $$ \frac{b^3}{a^3}=c $$ Hence, option (3) is correct.

Hence, the correct answer is option: 'C'.

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