| If \[{{x}^{1/3}}+{{y}^{1/3}}-{{z}^{1/3}}=0,\]then \[\{{{(x+y+z)}^{3}}+27xyz\}\]is equal to [SSC (CPO) 2007] |
A) \[-1\]
B) \[1\]
C) \[0\]
D) \[27\]
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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