The total number of distinct x \in R for which \left|\right. \textrm{ } x & x^{2} & 1 + x^{3} \\ 2 x & 4 x^{2} & 1 + 8 x^{3} \\ 3 x & 9 x^{2} & 1 + 27 x^{3} \textrm{ } \left|\right. = 10 is

Engineering

Mathematics

Properties of Determinant

Question

The total number of distinct x $\in$ R for which $| \begin{matrix} x & x^{2} & 1 + x^{3} \\ 2 x & 4 x^{2} & 1 + 8 x^{3} \\ 3 x & 9 x^{2} & 1 + 27 x^{3} \end{matrix} |$ = 10 is

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$∵ | \begin{matrix} x & x^{2} & 1 + x^{3} \\ 2 x & 4 x^{2} & 1 + 8 x^{3} \\ 3 x & 9 x^{2} & 1 + 27 x^{3} \end{matrix} | = 10$

$\Rightarrow | \begin{matrix} x & x^{2} & 1 \\ 2 x & 4 x^{2} & 1 \\ 3 x & 9 x^{2} & 1 \end{matrix} | + | \begin{matrix} x & x^{2} & x^{3} \\ 2 x & 4 x^{2} & 8 x^{3} \\ 3 x & 9 x^{2} & 27 x^{3} \end{matrix} | = 10$

$\Rightarrow x^{3} | \begin{matrix} 1 & 1 & 1 \\ 2 & 4 & 1 \\ 3 & 9 & 1 \end{matrix} | + x^{6} | \begin{matrix} 1 & 1 & 1 \\ 2 & 4 & 8 \\ 3 & 9 & 27 \end{matrix} | = 10$

⇒ x³ {1 (– 5) – 1 (– 1) + 1 (6)} + x⁶ {1 (36) – 1 (30) + 1 (6)} = 10

⇒ 2x³ + 12x⁶ = 10

⇒ 6x⁶ + x³ – 5 = 0 ⇒ (x³ + 1) (6x³ – 5) = 0

⇒ x³ = – 1 or $\frac{5}{6}$

$∴$ Two real solutions for x.