text: or
  
  # By using properties of determinants. Show that:(i) |1a a^2 1bb^2 1cc^2|=(a-b)(b-c)(c-a)(ii) |1 1 1a b c a^3b^3c^3|=(a-b)(b-c)(c-a)(a+b+c)
  
  Question from  Class 12  Chapter Determinants
  
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  Solution :
  (i) `L.H.S.= |[1,a,a^2],[1,b,b^2],[1,c,c^2]|`<br> Applying `R_1->R_1-R_2` and `R_2->R_2-R_3`<br> `= |[0,a-b,a^2-b^2],[0,b-c,b^2 - c^2],[1,c,c^2]|`<br> `= |[0,a-b,(a-b)(a+b)],[0,b-c,(b-c)(b+c)],[1,c,c^2]|`<br> `= (a-b)(b-c)|[0,1,(a+b)],[0,1,(b+c)],[1,c,c^2]|`<br> `= (a-b)(b-c)[b+c-a-b]`<br> `= (a-b)(b-c)(c-a) = R.H.S.`<br><br> (ii) `L.H.S. = |[1,1,1],[a,b,c],[a^3,b^3,c^3]|`<br> Applying `C_1->C_1- C_2 and C_2->C_2-C_3`<br> `=|[0,0,1],[a-b,b-c,c],[a^3-b^3,b^3-c^3,c^3]|`<br> `=|[0,0,1],[a-b,b-c,c],[(a-b)(a^2+ab+b^2),(b-c)(b^2+bc+c^2),c^3]|`<br> `=(a-b)(b-c)|[0,0,1],[1,1,c],[a^2+ab+b^2,b^2+bc+c^2,c^3]|`<br> `=(a-b)(b-c)[b^2+bc+c^2 - a^2- ab-b^2]`<br> `=(a-b)(b-c)[bc - ab +c^2 - a^2]`<br> `=(a-b)(b-c)[b(c-a)+(c+a)(c-a)]`<br> `=(a-b)(b-c)(c-a)(a+b+c) = R.H.S.`<br>
  
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