Question

If $ a, b $ and $ c $ are the roots of equation $ x^3 - px^2 + qx - r = 0 $ , then the value of $ \frac{1}{a^2} + \frac{1}{b^2} + \frac {1}{c^2} $ is

• A. $ \frac {p^2-2qr}{r^2} $
• B. $ \frac {q^2-2pr}{r^2} $
• C. $ \frac {r^2-2pq}{q^2} $
• D. $ \frac {r^2-2pq}{p^2} $

Answer 1

Answer: (B)

Since, $a, b$ and $c$ are the roots of equation $x^{3}-p x^{2}+q x-r=0$
$\therefore a+b+c=p, a b+b c+c a=q, a b c=r$
Now, $\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}=\frac{a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}}{(a b c)^{2}}$
$=\frac{(a b)^{2}+(b c)^{2}+(c a)^{2}}{(a b c)^{2}}$
$=\frac{(a b+b c+c a)^{2}-2\left(a b^{2} c+b c^{2} a+c a^{2} b\right)}{(a b c)^{2}}$
$=\frac{(a b+b c+c a)^{2}-2 a b c(a+b+c)}{(a b c)^{2}}$
$=\frac{q^{2}-2 r p}{r^{2}}$
$=\frac{q^{2}-2 p r}{r^{2}}$