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if x+y+z = 0, then show that x3+ y3+ z3= 3xyz.

We know that x3 + y3 + z3 - 3xyz = (x + y + z) (x2 + y2 + z2 - xy - yz - zx)

Now if x + y + z = 0, then......

x3 + y3 + z3 - 3xyz = ( 0 ) (x2 + y2 + z2 - xy - yz - zx)  . . . . . . .... ..... ... .. . . .. . .....    [ subsituting the value of x + y + z ]

x3 + y3 + z3 - 3xyz = 0    ......................  . . ..  . .. .  . . .  . . . . . .. .  .. . . .... . .    [ as any thing multiplied by 0 is 0 ]

x3 + y3 + z3 = 3xyz .      . .. .... .. . .  .. .  . . ... .. [ transporting -3xyz form L.H.S to R.H.S makeing the term positive ]

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 By using x3+y3+z3- 3xy =1/2 ( x+y+z) [(x-y)2 + (y-z)2 +(z-x)2] 

x3+ y3+ z3- 3xy = 1/2 (0) [ ( x-y)2+ (y-z)2 + (z-x)2] [ sustitute 0 in the place of x+y+z]

                    =  (0) [ Because 0 * something is 0]

 x3+ z3+ z3- 3xy= 0 [ take 3xy to the R.H.S]

therefore, x3+ y3+ z3 = 3xy

Hence proved.  

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x+y+z=0

x+y=(-z)  [1]   {transporting z to RHS}

cubing on both sides

(x+y)3=(-z)3

x3+y3+3xy(x+y)=-z3

x3+y3+z3=-3xy(x+y)

and, x+y=-z

putting in the value of x+y

x3+y3 +z3 =-3xy(-z)

therefore, x3+y3 +z3 =3xyz  hence proved

 

 

 

 

 

 

 

 


 

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