P.T [-bc b2+bc c2+bc]

[a2+ac -ac c2+ac]=(ab+bc+ca)3

[a2+ab b2+ab -ab]

Dear Linta

     |-bc  b²+bc  c²+bc |

     |a²+ac   -ac  c²+ac|

      |a²+ab  b²+ab  -ab|

multiple a in R1  b in R2 and c in R3

and then take common a from C1 ,b from C2 and c from C3

determinat will become

      |-bc   ab+ac   ac+ba |

     |ab+bc   -ac   bc+ab|

      |ac+cb  bc+ac  -ab|

nor R1 -> R1 + R2 +R3

 and take common (ab+bc+ca)

 determinat will become

                        |1           1         1 |

    (ab+bc+ca)    |ab+bc   -ac   bc+ab|

                        |ac+cb  bc+ac  -ab |

  now C1->C1 -c3

  and C2 - > C2-C3

and open determinat

        = (ab+bc+ca)³

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