If x13x^\frac{1}{3}x31 - y13y^\frac{1}{3}y31 = z13z^\frac{1}{3}z31, then find the value of (x - y - z)3 - 27xyz. A1 B27 C-1 D0 Show answer & explanationCorrect answer: DExplanationThe correct answer is option 4 i.e. 0. x1/3−y1/3=z1/3⇒x1/3=y1/3+z1/3a=x1/3, b=y1/3, c=z1/3⇒a=b+c(x−y−z)3=(a3−b3−c3)3 a3−b3−c3=3abc ( a=b+c)⇒(x−y−z)3=(3abc)3=27a3b3c3=27xyz(x−y−z)3−27xyz=0x^{1/3} - y^{1/3} = z^{1/3} \\ \Rightarrow x^{1/3} = y^{1/3} + z^{1/3} \\[6pt] a = x^{1/3},\ b = y^{1/3},\ c = z^{1/3} \\ \Rightarrow a = b + c (x - y - z)^3 = (a^3 - b^3 - c^3)^3 \\ \text{ } a^3 - b^3 - c^3 = 3abc \text{ ( } a = b + c) \\ \Rightarrow (x - y - z)^3 = (3abc)^3 = 27a^3b^3c^3 = 27xyz \\[6pt] \text (x - y - z)^3 - 27xyz = 0x1/3−y1/3=z1/3⇒x1/3=y1/3+z1/3a=x1/3, b=y1/3, c=z1/3⇒a=b+c(x−y−z)3=(a3−b3−c3)3 a3−b3−c3=3abc ( a=b+c)⇒(x−y−z)3=(3abc)3=27a3b3c3=27xyz(x−y−z)3−27xyz=0