Question

If xpower 1/3 +y power1/3+ zpower1/3=0 then find the value of x+y+z power cube

Answer 1

Answer:

$(x+y+z)^3=27xyz$

Step-by-step explanation:

Formula used:

If $x+y+z=0$ then $x^3+y^3+z^3=3xyz$

Given:

$x^{\frac{1}{3}}+y^{\frac{1}{3}}+z^{\frac{1}{3}}=0$

Using the above formula,

$(x^{\frac{1}{3}})^3+(y^{\frac{1}{3}})^3+(z^{\frac{1}{3}})^3=3(x^{\frac{1}{3}}y^{\frac{1}{3}}z^{\frac{1}{3}})$

$\implies\:x+y+z=3(xyz)^{\frac{1}{3}}$

Now,

$(x+y+z)^3$

$=(3(xyz)^{\frac{1}{3}})^3$

$=3^3((xyz)^{\frac{1}{3}})^3$

$=27xyz$