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Mathematics

If x13+y13+z13x^\frac{1}{3} + y^\frac{1}{3} + z^\frac{1}{3} = 0, then

  1. x3 + y3 + z3 = 0

  2. x3 + y3 + z3 = 27xyz

  3. (x + y + z)3 = 27xyz

  4. x + y + z = 3xyz

Factorisation

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Answer

If, x13+y13+z13x^\frac{1}{3} + y^\frac{1}{3} + z^\frac{1}{3} = 0

If, a + b + c = 0, then a3 + b3 + c3 = 3abc. 

Here, a = x13x^\frac{1}{3}, b = y13y^\frac{1}{3} and z = z13z^\frac{1}{3}

So,

⇒(x13)3+(y13)3+(z13)3=3(x13)(y13)(z13)⇒x33+y33+z33=3(xyz)13⇒x+y+z=3(xyz)13\Rightarrow (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})\\[1em] \Rightarrow x^\frac{3}{3} + y^\frac{3}{3} + z^\frac{3}{3} = 3(xyz)^\frac{1}{3}\\[1em] \Rightarrow x + y + z = 3(xyz)^\frac{1}{3}

Cubing both sides we get :

⇒(x+y+z)3=[3(xyz)13]3⇒(x+y+z)3=33.(xyz)33⇒(x+y+z)3=27xyz\Rightarrow (x + y + z)^3 = [3(xyz)^\frac{1}{3}]^3\\[1em] \Rightarrow (x + y + z)^3 = 3^3.(xyz)^\frac{3}{3}\\[1em] \Rightarrow (x + y + z)^3 = 27xyz

Hence, option 3 is the correct option.

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