Question

show that |1 a a³| |1 b b³| |1 c c³|=(a-b)(b-c)(c-a)​

Answer 1

Answer:

Given :   \begin{gathered}\left[\begin{array}{ccc}1&a&a^3\\1&b&b^3\\1&c&c^3\end{array}\right]\end{gathered}⎣⎢⎡111abca3b3c3⎦⎥⎤

To find : Prove that  given = (a - b)(b-c)(c-a)(a + b + )

Solution:

\begin{gathered}\left[\begin{array}{ccc}1&a&a^3\\1&b&b^3\\1&c&c^3\end{array}\right] =(a - b)(b-c)(c-a)(a + b + )\end{gathered}⎣⎢⎡111abca3b3c3⎦⎥⎤=(a−b)(b−c)(c−a)(a+b+)

LHS

= \begin{gathered}\left[\begin{array}{ccc}1&a&a^3\\1&b&b^3\\1&c&c^3\end{array}\right]\end{gathered}⎣⎢⎡111abca3b3c3⎦⎥⎤

= bc³ - cb³   - a(c³ - b³) + a³(c - b)

= bc(c² - b²) - a(c - b)(b² + c² + bc) +  a³(c - b)

= bc(c + b)(c - b)  - a(c - b)(b² + c² + bc) +  a³(c - b)

= (c - b) ( bc(c + b) - a(b² + c² + bc)  + a³ )

=  (c - b) ( bc² + b²c - ab² - ac² - abc   + a³ )

=  (c - b)(bc² - ac²  + b²c  - abc  + a³  - ab²)

= (c - b)(c²(b - a)  + bc(b  - a)  + a(a²  - b²))

= (c - b)(c²(b - a)  + bc(b  - a)  + a(a+b)(a - b))

=  (c - b)(b - a)( c²  + bc  - a(a + b))

= (b - c)(a - b) (c² + bc - a² - ab)

= (b - c)(a - b) (c²  - a² + bc - ab)

= (b - c)(a - b) ((c + a)(c - a)  + b(c - a))

= (b - c)(a - b) ( (c - a)  ((c + a) + b))

= (a - b)(b-c)(c - a)(a + b + c)

= RHS