Answers

2014-09-29T20:28:18+05:30
(a+b+c)^2= (a+b+c)(a+b+c)=a(a+b+c)+b(a+b+c)+c(a+b+c)
⇒a^2+ab+ac+ab+b^2+bc+ac+bc+c^2
⇒a^2+b^2+c^2+2ab+2bc+2ca
Hence proved

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((a+b)+(c))^2=(a+b)^2+2(a+b)(c)+(c)^2=a^2+2ab+b^2+2ac+2bc+c^2
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2014-09-29T20:35:33+05:30
Let  (x + y ) = t then,
 
(x + y + z) = (t +z) ²
                 = ( t² + 2tz + t²)------------------ ( using identity I)
                   = (x +y )² + 2(x+y)z + z²----------( sub the value t )
                     = x² + 2xy + y² = 2xz +2yz + z²----------(identity I)
                 x²+y²+z²+2xy+2xy+ 2yz+2 zx
                  
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