If x^{1/3} + y^{1/3} = z^{1/3}, then (x + y - | vedclass

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(A) Given: $x^{1/3} + y^{1/3} = z^{1/3}$ ......$(1)$Cube both sides of the equation:$(x^{1/3} + y^{1/3})^3 = (z^{1/3})^3$Using the identity $(a + b)^3

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(A) Given: $x^{1/3} + y^{1/3} = z^{1/3}$ ......$(1)$Cube both sides of the equation:$(x^{1/3} + y^{1/3})^3 = (z^{1/3})^3$Using the identity $(a + b)^3 = a^3 + b^3 + 3ab(a + b)$:$x + y + 3x^{1/3}y^{1/3}(x^{1/3} + y^{1/3}) = z$Substitute $x^{1/3} + y^{1/3} = z^{1/3}$ from equation $(1)$:$x + y + 3x^{1/3}y^{1/3}z^{1/3} = z$Rearrange the terms:$x + y - z = -3x^{1/3}y^{1/3}z^{1/3}$Cube both sides again:$(x + y - z)^3 = (-3x^{1/3}y^{1/3}z^{1/3})^3$$(x + y - z)^3 = -27xyz$Therefore:$(x + y - z)^3 + 27xyz = 0$

If $x^{1/3} + y^{1/3} = z^{1/3},$ then $(x + y - z)^3 + 27xyz$ is equal to

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