Question

Please sove it i m unable to solve it
can any one help me

Answer 1

▶hey mate◀

here's ur answer ⬇⬇❤

we know that

if a+b+c =0 then,

a3+b3+c3=3abc

$so \: if \: x { }^{ \frac{1}{3} } + y {}^{ \frac{1}{3} } + z {}^{ \frac{1}{3} } = 0 \\ then \: x + y + z = 3( x {}^{ \frac{1}{3} } + y {}^{ \frac{1}{3} } + z {}^{ \frac{1}{3} }) \\ \\ so \: (x + y + z) {}^{3} = (3x {}^{ \frac{1}{3} } + y {}^{ \frac{1}{3} } + z {}^{ \frac{1}{3} }) {}^{3} \\ \\ therefore.. (x + y + z) {}^{3} = 27xyz$

hope it helps :-)

plzz mark it as branliest ..

#thnk u

Answer 2

x^1/3 +y^1/3 +z^1/3 =0 ( if a+b+c=0
x+y+z=3(xyz)^1/3 a^3 +b^3 +c^3=3abc)

Now
(x+y+z)^3
=(3×(xyz)^1/3)^3
=3^3 ×(xyz)^3×1/3
=27xyz Ans