Question

plzz answer no.29 in the following image through indices

Answer 1

x⅓ + y⅓ + z⅓ = 0

let x⅓ = a
y⅓ = b
z⅓ = c

then, a + b + c = 0
we know,
a³ + b³ + c³ = 3abc
when, a + b + c = 0

now,
a³ + b³ + c³ = 3abc

(x⅓)³ + (y⅓)³ + (z⅓)³ = 3(xyz)⅓

x + y + z = 3(xyz)⅓

take cube both sides,

(x + y + z)³ = (3)³(xyz)⅓ ×3 = 27xyz

hence proved//

Answer 2

hi friend,

given x⅓+y⅓+z⅓ =0

let x⅓=a

y⅓=b

z⅓=c

<>we know that a³+b³+c³=3abc
if a+b+c=0

→(x⅓)³+(y⅓)³+(z⅓)³=3x⅓y⅓z⅓

→x+y+z=3x⅓y⅓z⅓

cubing on both sides, we get

→(x+y+z)³=27xyz

I hope this will help u ;)