Question

[tex] prove that \left|\begin{array}{ccc} 1&1&1 \\a²&b²&c²\\a³&b³&c³\end{array}\right| = (a-b)(b-c)(c-a)(ab+bc+ca) [\tex]

Answer 1

Step-by-step explanation:

To prove

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

$C1 -> C1 - C2 \\ \\ C2 -> C2 - C3$

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

As we know that

${a}^{2} - {b}^{2} = (a + b)(a - b) \\ \\ {a}^{3} - {b}^{3} = (a - b)( {a}^{2} + ab + {b}^{2} )$

$\left|\begin{array}{ccc} 0&0&1 \\(a-b)(a+b)&(b-c)(b+c)&c^2\\(a-b)(a^2+ab+b^2)&(b-c)(b^2+bc+c^2)&c^3\end{array}\right|$

Take (a-b) from C1 and (b-c) from C2

$(a-b)(b-c)\left|\begin{array}{ccc} 0&0&1 \\(a+b)&(b+c)&c^2\\(a^2+ab+b^2)&(b^2+bc+c^2)&c^3\end{array}\right|$

C1 -> C1-C2

$(a-b)(b-c)\left|\begin{array}{ccc} 0&0&1 \\a+b-b-c&(b+c)&c^2\\a^2+ab+b^2-b^2-bc-c^2&(b^2+bc+c^2)&c^3\end{array}\right|$

$(a-b)(b-c)\left|\begin{array}{ccc} 0&0&1 \\a-c&(b+c)&c^2\\a^2-c^2+ab-bc&(b^2+bc+c^2)&c^3\end{array}\right|$

$(a-b)(b-c)\left|\begin{array}{ccc} 0&0&1 \\a-c&(b+c)&c^2\\a^2-c^2+ab-bc&(b^2+bc+c^2)&c^3\end{array}\right|$

$(a-b)(b-c)\left|\begin{array}{ccc} 0&0&1 \\a-c&(b+c)&c^2\\(a-c)(a+b+c)&(b^2+bc+c^2)&c^3\end{array}\right|$

Take common (a-c) from C1

$(a-b)(b-c)(a-c)\left|\begin{array}{ccc} 0&0&1 \\1&(b+c)&c^2\\(a+b+c)&(b^2+bc+c^2)&c^3\end{array}\right|$

Expand the determinant along R1

${b}^{2} + bc + {c}^{2} - (b + c)(a + b + c) \\ \\ {b}^{2} + bc + {c}^{2} - ab - {b}^{2} - bc - ac - bc - {c}^{2} ) \\ \\ - (ab + bc + ac)$

So LHS becomes

$= > - (a-b)(b-c)(a-c)( ab + bc + ac) \\ \\ = > (a-b)(b-c)(c - a)( ab + bc + ac) \\ \\ =RHS$

Hope it helps you.