Question

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

Answer 1

Answer:

we know that

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

it is given x+y+z=0

0(x²+y²-xy-zy-zx)=x³+y³+z³-3xyz

0=x³+y³z³-3xyz

3xyz=x³+y³+z³

Answer 2

ɢɪᴠᴇɴ:

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

sᴏʟᴜᴛɪᴏɴ:

x + y + z = 0 ⇨ x + y = -z

➙ (x + y)³ = ( -z)³

➙ x³ + y³ + 3xy(x + y) = (-z)³

➙x³ + y³ + 3xy( -z ) = -z³ [∵ (x+y) = -z ]

➙ x³ + y³ - 3xyz = -z³

➙ x³ + y³ + z³ = 3xyz

Hence,

( x + y + z) = 0

➙ (x³ + y³ + z³) = 3xyz

Hence Proved !!!