Question

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

| 1 c²+ab c³|
​

Answer 1

Answer:

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Step-by-step explanation:

 

LHS =  

C₁ →  C₁ - C₂    & C₂ →  C₂ - C₃    

=  

=  

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

=  

C₂ →  C₂ - C₁  

=  

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

=  

Taking common  c - a   from C₂

=  

= (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