Prove that determinant ((1,1,1),(a,b,c),(a^2 ,b^2 ,c^2 ))= (a−b)(b−c)(c−a)


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Question Number 175110 by nadovic last updated on 19/Aug/22

$Prove that \| \begin{matrix} 1 & 1 & 1 \\ a & b & c \\ a^{2} & b^{2} & c^{2} \end{matrix} \| = (a - b) (b - c) (c - a)$
Commented by kaivan.ahmadi last updated on 19/Aug/22

$\| \begin{matrix} 1 1 1 \\ a b c \\ a^{2} b^{2} c^{2} \end{matrix} \| \| \begin{matrix} 1 1 1 \\ a b c \\ a^{2} b^{2} c^{2} \end{matrix} \| =$ $({b c}^{2} + {c a}^{2} + {a b}^{2}) - ({b a}^{2} + {a c}^{2} + {c b}^{2})$ $= c (b c + a^{2} - a c - b^{2}) - a b (a - b)$ $= c (c (b - a) + (a - b) (a + b)) - a b (a - b)$ $= (a - b) (- c^{2} + a c + b c - a b)$ $= (a - b) (c (a - c) + b (c - a))$ $= (a - b) ((c - a) (b - c)) =$ $(a - b) (b - c) (c - a)$ $n o t e t h a t t h i s i s a V a n d e r m o n d e$ $d e t e r m i n a n t .$ $I n g e n e r a l c a s e$ $\| \begin{matrix} 1 1 . . . 1 \\ a_{1} a_{2} . . . a_{n} \\ ⋮ ⋮ ⋮ \\ a_{1}^{n - 1} . . . a_{n}^{n - 1} \end{matrix} \| = \prod_{1 ⩽ i < j ⩽ n} (a_{j} - a_{i})$
Commented by Tawa11 last updated on 19/Aug/22

$Great sir .$
Commented by puissant last updated on 21/Aug/22

$V a n d e r m o n d e m e t h o d$
	Answered by Rasheed.Sindhi last updated on 19/Aug/22
	$Prove that determinant ((1,1,1),(a,b,c),(a^2 ,b^2 ,c^2 ))= (a−b)(b−c)(c−a) ∣ (/)Through properties(/) ∣ ^(−) = determinant ((1,( 0),( 0)),(a,(b−a),(c−a)),(a^2 ,(b^2 −a^2 ),(c^2 −a^2 )))C2−C1_(C3−C1) =(b−a)(c−a) determinant ((1,( 0),( 0)),(a,( 1),( 1)),(a^2 ,(b+a),(c+a))) =(b−a)(c−a)∙1{1(c+a)−1(b+a)} =(b−a)(c−a)(c−b) =(a−b)(b−c)(c−a)$
	$Prove that \| \begin{matrix} 1 & 1 & 1 \\ a & b & c \\ a^{2} & b^{2} & c^{2} \end{matrix} \| = (a - b) (b - c) (c - a)$ $\overset{―}{\underset{―}{∣} \underset{―}{\frac{}{} Through properties \frac{}{} ∣}}$ $= \| \begin{matrix} 1 & 0 & 0 \\ a & b - a & c - a \\ a^{2} & b^{2} - a^{2} & c^{2} - a^{2} \end{matrix} \| \underset{C 3 - C 1}{C 2 - C 1}$ $= (b - a) (c - a) \| \begin{matrix} 1 & 0 & 0 \\ a & 1 & 1 \\ a^{2} & b + a & c + a \end{matrix} \|$ $= (b - a) (c - a) \cdot 1 {1 (c + a) - 1 (b + a)}$ $= (b - a) (c - a) (c - b)$ $= (a - b) (b - c) (c - a)$
	Commented by puissant last updated on 21/Aug/22

	$G r e a t$
	Commented by Rasheed.Sindhi last updated on 21/Aug/22

	$T h a n k s s i r!$
	Answered by Rajpurohith last updated on 10/Jun/23

	$A p p l y t h e o p e r a t i o n C_{1} \to C_{1} - C_{2} a n d C_{2} \to C_{2} - C_{3}$ $T h e d e t e r m i n a n t r e d u c e s t o,$ $\| \begin{matrix} 0 0 1 \\ a - b b - c c \\ a^{2} - b^{2} b^{2} - c^{2} c^{2} \end{matrix} \| = (a - b) (b - c) \| \begin{matrix} 0 0 1 \\ 1 1 c \\ a + b b + c c^{2} \end{matrix} \|$ $T h i s i s b y t a k i n g t h e c o m m o n f a c t o r s f r o m C_{1} & C_{2}$ $a n d o p e r a t e C_{1} \to C_{1} - C_{2}$ $= (a - b) (b - c) \| \begin{matrix} 0 0 1 \\ 0 1 c \\ a - c b + c c^{2} \end{matrix} \|$ $e x p a n d i n g a l o n g t h e f i r s t r o w,$ $= (a - b) (b - c) (c - a)$
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