If a, b, c are distinct and $(a, a^{2}, a^{3} + 1)$ , $(b, b^{2}, b^{3} + 1)$ , $(c, c^{2}, c^{3} + 1)$ are linearly dependent, then value of abc is

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Solution

The correct option is A $- 1$
Linearly dependent when
$∣ ∣ ∣ ∣ ∣ \begin{matrix} a & a^{2} & a^{3} + 1 b & b^{2} & b^{3} + 1 c & c^{2} & c^{3} + 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0$
Applying $R_{2} \to R_{2} - R_{1}, R_{3} \to R_{3} - R_{1}$
$∣ ∣ ∣ ∣ ∣ \begin{matrix} a & a^{2} & a^{3} + 1 b - a & b^{2} - a^{2} & b^{3} - a^{3} c - a & c^{2} - a^{2} & c^{3} - a^{3} \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0 \Rightarrow (b - a) (c - a) ∣ ∣ ∣ ∣ ∣ \begin{matrix} a & a^{2} & a^{3} + 1 0 & b + a & b^{2} + a^{2} + a b 0 & c + a & c^{2} + a^{2} + a c \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0 \Rightarrow (b - a) (c - a) (a - b) (a b c + 1) = 0$
Thus $a b c = - 1$

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If a, b, c are in G.P., prove that:

(i) $a (b^{2} + c^{2}) = c (a^{2} + b^{2})$

(ii) $a^{2} b^{2} c^{2} (\frac{1}{a^{3}} + \frac{1}{b^{3}} + \frac{1}{c^{3}}) = a^{3} + b^{3} + c^{3}$

(iii) $\frac{(a + b + c)^{2}}{a^{2} + b^{2} + c^{2}} = \frac{a + b + c}{a - b + c}$

(iv) $\frac{1}{a^{2} - b^{2}} + \frac{1}{b^{2}} = \frac{1}{b^{2} - c^{2}}$

(v) $(a + 2 b + 2 c) (a - 2 b + 2 c) = a^{2} + 4 c^{2}$ .

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