Question

If x+y+z=0,show that x³+y³+z³=3xyz​

Answer 1

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Step-by-step explanation:

$x + y + z = 0$

$x + y = - z$

${(x + y)}^{3} = {( - z)}^{3}$

${x}^{3} + 3xy(x + y) + {y}^{3} = - {z}^{3}$

${x}^{3} + {y}^{3} + {z}^{3} = - 3xy(x + y)$

Now,

$x + y + z = 0 \\ x + y = - z$

So,

${x}^{3} + {y}^{3} + {z}^{3} = - 3xy \times ( - z)$

${x}^{3} + {y}^{3} + {z}^{3} = 3xyz$

Hence proved.

Answer 2

Given

x+y+z=0

To prove

x³+y³+z³=3xyz

$\sf\huge {\underline{\underline{{Solution}}}}$

x³+y³+z³=3xyz

shifts 3xyz to left side

=> x³+y³+z³-3xyz

It can also be written as

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

Now ,put x+y+z=0

=>x³+y³+z³-3xyz= 0(x²+y²+z²-xy-yz-zx)

Now,on the right side the whole term be zero

=>x³+y³+z³-3xyz=0

=> x³+y³+z³= 3xyz( proved )

Alternative Method

x+y+z=0

x+y= -z------(1)

Now , cubing both sides

=>(x+y)³=(-z)³

Identity : (a+b)³= a³+b³+3a²+3ab²

=>x³+y³+3x²y+3xy²= -z³

Taking 3xy common

=>x³+y³+3xy( x+y)= -z³

From Equation 1 put z in place of x+y

=>x³+y³+3xy(-z)= -z³

=> x³+y³-3xyz= -z³

Now shifts -z³ to left side and -3xyz to right side

=>x³+y³+z³= 3xyz( proved )