Question

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

Answer 1

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

To find : To be Proved

Solution:

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

LHS =  $\left[\begin{array}{ccc}1&1&1\\a^2&b^2&c^2\\a^3&b^3&c^3\end{array}\right]$

C₁ →  C₁ - C₂    & C₂ →  C₂ - C₃    

= $\left[\begin{array}{ccc}0&0&1\\a^2-b^2&b^2-c^2&c^2\\a^3-b^3&b^3-c^3&c^3\end{array}\right]$

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

Taking (a - b) common from C₁  & b -c from C₂

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

C₂ →  C₂ - C₁  

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

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

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

Taking common  c - a   from C₂

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

= (a - b)(b - c)(c-a) (  0 - 0  +  1 ( a +b)(a+b+c) - (a² + b² + ab) * 1 )

=(a - b)(b - c)(c-a) (  a² + ab  + ac +  ab + b² + bc- (a² + b² + ab) * 1 )

= (a - b)(b - c)(c-a)  (ab + bc + ca)

= RHS

QED

Hence Proved

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