Question

if x+y+z=0. show that x^3 + y^3 + z^3= 3xyz​

Answer 1

Solution :-

x + y + z = 0

⇒ x + y = - z --- eq(1)

Cubing on both sides

⇒ (x + y)³ = (-z)³

⇒ x³ + y³ + 3xy(x + y) = - z³

[ Because (x + y)³ = x³ + y³ + 3xy(x + y) ]

⇒ x³ + y³ + 3xy(-z) = - z³

[ From eq(1) ]

⇒ x³ + y³ - 3xyz = - z³

⇒ x³ + y³ + z³ = 3xyz

Hence shown.

Answer 2

$\bf Given : \\ x + y + z = 0$

$\bf To \: prove : \\ {x}^{3} + {y}^{3} + {z}^{3} = 3xyz$

$\bf Solution : \\ \Rightarrow x + y + z = 0 \\ \\ \Rightarrow x + y = - z \\ \\ \bf cubing \: both \: side \\ \\ \Rightarrow {(x + y)}^{3} = { - z}^{3} \\ \\ \Rightarrow {x}^{3} + {y}^{3} + 3xy(x + y) = { - z}^{3} \\ \\ \bf replace \: x + y \: with \: - z \\ \\ \Rightarrow{x}^{3} + {y}^{3} + 3xy( - z) = { - z}^{3} \\ \\ \Rightarrow {x}^{3} + {y}^{3} - 3xyz = - {z}^{3} \\ \\ \bf \boxed{\Rightarrow {x}^{ 3 } + {y}^{ 3} + {z}^{3} = 3xyz}$

$\bf Hence \: proved$