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

Please feel free to post as many doubts on our discussion forum as you can.
If you find any question Difficult to understand - post it here and we will get you
the answer and detailed  solution very  quickly.

 We are all IITians and here to help you in your IIT JEE preparation.

All the best.
 
Regards,
Askiitians Experts
Badiuddin