Question

These is the question of class 12 maths​

Answer 1

Question: To prove that, |1 a² a³, 1 b² b³, 1 c² c³| = (a - b)(b - c)(c - a)(ab + bc + ca)

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$\Longrightarrow \left|\begin{array}{ccc}\sf 1 & \sf a^2 & \sf a^3 \\ \sf 1 & \sf b^2 & \sf b^3 \\\sf 1 & \sf c^2 & \sf c^3\end{array}\right| = \sf (a - b)(b - c)(c - a)(ab + bc + ca)$

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Taking the LHS we get;

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$\Longrightarrow \left|\begin{array}{ccc}\sf 1 & \sf a^2 & \sf a^3 \\ \sf 1 & \sf b^2 & \sf b^3 \\\sf 1 & \sf c^2 & \sf c^3\end{array}\right|$  

 

Let R1 ➝ R1 - R2

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$\Longrightarrow \left|\begin{array}{ccc}\sf 1 - 1 & \sf a^2 - b^2 & \sf a^3 - b^3\\ \sf 1 & \sf b^2 & \sf b^3 \\\sf 1 & \sf c^2 & \sf c^3\end{array}\right|$

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Let R2 ➝ R2 - R3

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$\Longrightarrow \left|\begin{array}{ccc}\sf 0 & \sf a^2 - b^2 & \sf a^3 - b^3\\ \sf 1 - 1 & \sf b^2 - c^2 & \sf b^3 - c^3 \\\sf 1 & \sf c^2 & \sf c^3\end{array}\right|$

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$\Longrightarrow \left|\begin{array}{ccc}\sf 0 & \sf a^2 - b^2 & \sf a^3 - b^3\\ \sf 0 & \sf b^2 - c^2 & \sf b^3 - c^3 \\\sf 1 & \sf c^2 & \sf c^3\end{array}\right|$

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On expanding the determinant along C1 (first column) we get:

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$\Delta = \sf 0\left|\begin{array}{ccc} \sf b^2 - c^2 & \sf b^3 - c^3 \\ \sf c^2 & \sf c^3\end{array}\right| - 0\left|\begin{array}{ccc} \sf a^2 - b^2 & \sf a^3 - b^3 \\ \sf c^2 & \sf c^3 \end{array}\right| + 1\left|\begin{array}{ccc} \sf a^2 - b^2 & \sf a^3 - b^3\\ \sf b^2 - c^2 & \sf b^3 - c^3\end{array}\right|$

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$\Delta = \sf 0 - 0 + 1\left|\begin{array}{ccc} \sf a^2 - b^2 & \\ \sf b^2 - c^2 \end{array} \! \! \! \! \! \! \nearrow \! \! \! \! \! \! \searrow \begin{array}{ccc} \sf a^3 - b^3\\ \sf b^3 - c^3\end{array}\right|$

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$\sf \Delta = 1 \Big[(a^2 - b^2)(b^3 - c^3) - \big[(a^3 - b^3)(b^2 - c^2)\big]\Big]$

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Using the following identities we get:

• a² - b² = (a + b)(a - b)
• a³ - b³ = (a - b)(a² + ab + b²)

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$\sf \Delta = (a + b)(a - b)(b - c) (b^2 + bc + c^2) - (a - b) (a^2 + ab + b^2)(b^2 - c^2)$

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(a - b) is a common factor in both the subtrahend, so let's take it out as a common factor & expand (b² - c²).

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$\sf \Delta = (a - b)\Big[(a + b)(b - c) (b^2 + bc + c^2) - (a^2 + ab + b^2)(b + c)(b - c)\Big]$

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(b - c) is a common factor in both the subtrahend, so let's take it out as a common factor

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$\sf \Delta = (a - b)(b - c)\Big[(a + b)(b^2 + bc + c^2) - (a^2 + ab + b^2)(b + c)\Big]$

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On opening the brackets in [] we get,

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$\Delta = \sf (a - b)(b - c)\Big[ab^2 + abc + ac^2 + b^3 + b^2c + bc^2 \\{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \sf - (ba^2 + ab^2 + b^3 + ca^2 + abc + cb^2)\Big]}$

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$\Delta = \sf (a - b)(b - c)\Big[\textsf{\textbf{ab}}^2 + \textsf{\textbf{abc}} + ac^2 + \textsf{\textbf{b}}^3 + \textsf{\textbf{b}}^2\textsf{\textbf{c}} + bc^2 \\{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ - ba^2 \textsf{\textbf{\ -\ ab}}^2 \textsf{\textbf{\ - b}}^3 - ca^2 \textsf{\textbf{\ - abc - cb}}^2 \Big]}$

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[The terms in bold have been cancelled]

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$\Delta = \sf (a - b)(b - c)\Big[ac^2 + bc^2 - ba^2 - ca^2 \Big]$

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On re-arranging the terms we get,

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$\Delta = \sf (a - b)(b - c)\Big[ac^2 - ca^2 + bc^2 - ba^2 \Big]$

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On taking out common factors we get,

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$\Delta = \sf (a - b)(b - c)\Big[ac(c - a) + b(c^2 - a^2) \Big]$

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On applying the identity a² - b² = (a + b)(a - b) we get,

[Where 'a' = c and 'b' = a]

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$\sf \Delta = (a - b)(b - c)\Big[ac(c - a) + b(c - a)(c + a) \Big]$

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(c - a) is common in both the subtrahend, on taking it out as a common factor we get,

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$\Delta = \sf (a - b)(b - c)(c - a)\Big[ac + b(c + a) \Big]$

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$\Delta = \sf (a - b)(b - c)(c - a)\Big[ac + bc + ba \Big]$

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On re-arranging the terms we get,

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$\underline{\underline{\bold\Delta = \sf \textsf{\textbf{(a - b)(b - c)(c - a)}}\big(\textsf{\textbf{ab + bc + ca}}\big)}}$

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LHS = RHS

Hence proved.