If (*x)^1/3 + (y)^1/3 + (z)^1/3= 3xyz, then*

*(a)x^3+y^3+z^3=0 (b)x+y+z=27xyz*

*(c)(x+y+z)^3=27xyz (d)x^3+y^3+z^3=27xyz*

Which option is correct? Please explain with steps.

### 1 Answer | Add Yours

You need to raise to cube both sides, such that:

`(x^(1/3) + y^(1/3) + z^(1/3))^3 = (3xyz)^3`

`(x^(1/3) + y^(1/3) + z^(1/3))^2*(x^(1/3) + y^(1/3) + z^(1/3)) = 27(xyz)^3`

`(x^(2/3) + y^(2/3) + z^(2/3) + 2(xy)^(1/3) + 2(xz)^(1/3) + 2(yz)^(1/3))(x^(1/3) + y^(1/3) + z^(1/3)) = 27(xyz)^3`

`x + y + z + x^(2/3)y^(1/3) + x^(2/3)z^(1/3) + x^(2/3)y^(1/3) + y^(2/3)z^(1/3) + x^(2/3)z^(1/3) + y^(2/3)z^(1/3) + 2x^(2/3)y^(1/3) + 2x^(1/3)y^(2/3) + 6(xyz)^(1/3) + 2y^(2/3)z^(1/3) + 2y^(1/3)z^(2/3) + 2x^(2/3)z^(1/3) + 2x^(1/3)z^(2/3) = 27(xyz)^3`

**Notice that only the option `x^3+y^3+z^3=0` check the given condition `x^(1/3) + y^(1/3) + z^(1/3) = 3xyz` , if `x=y=z=0` , hence, you may select the answer `a) x^3+y^3+z^3=0` . **

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