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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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Answer 1

Answer:

Answer:hhhiiiStep-by-step explanation: LHS = C₁ → C₁ - C₂ & C₂ → C₂ - C₃ = = Taking (a - b) common from C₁ & b -c from C₂ = C…

Answer 2

Answer:

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

Solving LHS

Expanding through C1,

$\triangle = 1\left|\begin{array}{cc}b^2 & b^3 \\ c^2 & c^3\end{array}\right| -1\left|\begin{array}{cc}a^2 & a^3 \\ c^2 & c^3\end{array}\right| +1\left|\begin{array}{cc}a^2 & a^3 \\ b^2 & b^3\end{array}\right|$

$\sf \bf :\longmapsto \triangle = b^2c^3 - b^3c^2 -a^2c^3 + a^3c^2$

Solving LHS:

$\sf \bf :\longmapsto (a-b)(b-c)(c-a)(ab+bc+ca)$

$\sf \bf :\longmapsto (ab-ac-b^2+bc)(c-a)(ab+bc+ca)$

$\sf \bf :\longmapsto (-ac^2 + bc^2 -b^2c -a^2b +a^2c +ab^2)(ab+bc+ca)$

$\sf \bf :\longmapsto -a^2bc^2+ab^2c^2-ab^3c$

$\sf \bf -a^3b^2+a^3bc+a^2b^3-abc^3+b^2c^3-b^3c^2$

$\sf \bf -a^2b^2c +a^2bc^2 + ab^3c -a^2c^3 -ab^2c^2$

$\sf \bf -a^3bc+a^3c^2+a^2b^2c+abc^3$

$\sf \bf :\longmapsto b^2c^3-b^3c^2-a^2c^3+a^3c^2+a^2b-a^3b^2$

= RHS

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