Question

if (x+y+z)=0,then prove that x^3+^3+z^3=3xyz​

Answer 1

Step-by-step explanation:

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

Cubing on both sides we get

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

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

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

$3xyz \: since \: x + y = - z$

Hence proved. Hope it helps you.

Answer 2

$\bf{\large{\underline{\underline{Correct \: Question:-}}}}$

If x + y + z = 0, then prove that x³ + y³ + z³ = 3xyz

$\bf{\large{\underline{\underline{GIVEN:-}}}}$

x + y + z = 0

$\bf{\large{\underline{\underline{REQUIRED \: TO \: PROVE:-}}}}$

x³ + y³ + z³ = 3xyz

$\bf{\large{\underline{\underline{PROOF:-}}}}$

$\tt x + y + z = 0$

Add - z on both the sides

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

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

$\tt x + y =- z.....(1)$

Cubing on both sides

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

We know that (a + b)³ = a³ + b³ + 3ab(a + b)

Here a = x, b = y

By substituting the values in the identity we have

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

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

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

[From. eq(1) i.e x + y = -z]

$\tt {x}^{3} + {y}^{3} - 3xyz = - {z}^{3}$

Add - 3xyz on both sides

$\tt {x}^{3} + {y}^{3} - 3xyz + 3xyz = - {z}^{3} + 3xyz$

$\tt {x}^{3} + {y}^{3} = - {z}^{3} + 3xyz$

Add z³ on both sides

$\tt {x}^{3} + {y}^{3} + {z}^{3} = - {z}^{3} + 3xyz + {z}^{3}$

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

Hence Proved