Using properties of determinants prove that -

(b+c)2....a2........a2

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

c2.....c2.......(a+b)2

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 :
https://www.meritnation.com/ask-answer/question/b-c-2-a-2-a-2-b-2-c-a-2-b-2-2abc-a-b-c-3/determinants/2386483

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Is t his in ncert??

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ummmmmmmmmmmmmm

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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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