Question

If 3x+y+z=0 then show that 27x^3+y^3+z^3=9xyz

Answer 1

3x+y+z=0=> 3x+y = -z On cubing on both sides, we get =>
${(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)$
${(3x + y)}^{3} = { - z}^{3} = > {(3x)}^{3} + {y}^{3} +$
$3(3x)(y)(3x + y) = > 27 {x}^{3} + {y}^{3} + {z}^{3} =$
$= - 9xy(3x + y) = > 27 {x}^{3} + {y}^{3} + {z}^{3}$
$so \: 27 {x}^{3} + {y}^{3} + {z}^{3} = 9xyz$
Hence showed .

Answer 2

3x+y+z=0
3x+y=-z
=(3x+y)³=-z³
= 27x³+y³+9xy(3x+y)=-z³
= 27x³+y³+9xy(-z)=-z³. [ (3x+y)=-z]
= 27x³+y³-9xyz=-z³
= 27x³ +y³ +z³ = 9xyz
Hence proved