Quantitative Aptitude Ques 1133

Question: If $ {x^{1/3}}+{y^{1/3}}-{z^{1/3}}=0, $ then $ {{{(x+y+z)}^{3}}+27xyz} $ is equal to

Options:

A) $ -1 $

B) $ 1 $

C) $ 0 $

D) $ 27 $

Show Answer

Answer:

Correct Answer: C

Solution:

  • $ {x^{1/3}}+{y^{1/3}}-{z^{1/3}}=0 $ $ {x^{1/3}}+{y^{1/3}}={z^{1/3}} $ On cubing both sides, we get $ {{({x^{1/3}}+{y^{1/3}})}^{3}}={z^{1/3\times 3}} $

$ \Rightarrow $ $ x+y+3{x^{1/3}}\cdot {y^{1/3}}({x^{1/3}}+{y^{1/3}})=z $ $ [\because {{(a+b)}^{3}}=a^{3}+b^{3}+3ab(a+b)] $

$ \Rightarrow $ $ x+y-z=3\cdot {x^{1/3}}\cdot {y^{1/3}}\cdot {z^{1/3}} $ … (i) Now, $ {{(x+y-z)}^{3}}+27xyz $ $ ={{(-3{x^{1/3}}\cdot {y^{1/3}}\cdot {z^{1/3}})}^{3}}+27xyz $ [from Eq. (i)] $ =-27xyz+27xyz=0 $

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