Using properties of determinants prove that -


b2.....(c+a)2......b2 =2abc(a+b+c)3


In this ques.. i just want to know tht after applying C1→ C1-C2, C2→ C2-C3

in this ques how can i take (a+b+c) common from C1 and C2.

 After applying the properties that is 
C1 = C‚Äč1 - C3  and C= C- C, we get 

 =  (b+c)2-a20a20(c+a)2-b2b2c2 -(a+b)2c2 -(a+b)2(a+b)2 = (a+b+c)(b+c-a)0a20(a+b+c)(c+a-b)b2(a+b+c)(c-a-b)(a+b+c)(c-a-b)(a+b)2so we can take (a+b+c) common from C1 and C2.=(a+b+c)2 (b+c-a)0a20(c+a-b)b2(c-a-b)(c-a-b)(a+b)2

Hope your doubt is cleared now.
Refer the following link :

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Is t his in ncert??

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yes.. thats why we applied the above mentioned properties ( to uncomplicate the matters) ..... then after making two zeros, we expand it.

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no.. no ..m asking how to take out (a+b+c) common from C1nd C2.

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