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in Determinants by (40.4k points)
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सिद्ध कीजिए कि  \(\begin{vmatrix}a^2&bc&ac+c^2\\a^2+ab & b^2&ac\\ab&b^2+bc&c^2\end{vmatrix}\) = 4a2b2c2.

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L.H.S. =  \(\begin{vmatrix}a^2&bc&ac+c^2\\a^2+ab & b^2&ac\\ab&b^2+bc&c^2\end{vmatrix}\) 

R1 के सापेक्ष विस्तार करने पर,

= 2abc2 [0 – 1{ac – b(a – c)} + 1{a(b + c) – ( – c) (b)]

= 2abc2 [- ac + ba – bc + ab + ac + bc]

= 2abc2 (2ab)

= 4a2b2c2

= R.H.S.
इति सिद्धम्।

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