Question

If \[\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}=0\] then find the value of \[{{\left( a+b+c \right)}^{3}}.\]
A. \[27abc\]
B. \[3abc\]
C. \[9abc\]
D. \[abc\]

Answer 1

Hint:Consider the values of the given expression as x, y, z for the terms given. Use the algebraic expansion of \[\left( {{x}^{3}}+{{y}^{3}}+{{z}^{3}} \right)\] to solve and find the expression for \[{{\left( a+b+c \right)}^{3}}.\]

Complete step-by-step answer:
Given is that \[\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}=0\]
Let us take the value of \[x=\sqrt[3]{a},y=\sqrt[3]{b},z=\sqrt[3]{c}\]
Here, we need to find the value of \[{{\left( a+b+c \right)}^{3}}.\]
We know the expression, \[\left( {{x}^{3}}+{{y}^{3}}+{{z}^{3}} \right)\].
\[{{x}^{3}}+{{y}^{3}}+{{z}^{3}}=\left( x+y+z \right)\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-xz \right)+3xyz\].
Here \[x+y+z=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\]
We have been given that \[x+y+z=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}=0.\]
So, \[x+y+z=0.\]
\[\begin{align}
  & {{x}^{3}}+{{y}^{3}}+{{z}^{3}}=\left( x+y+z \right)\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-xz \right)+3xyz \\
 & {{x}^{3}}+{{y}^{3}}+{{z}^{3}}=0\times \left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-xz \right)+3xyz \\
 & \Rightarrow {{x}^{3}}+{{y}^{3}}+{{z}^{3}}=3xyz.........(1) \\
\end{align}\]
We got an expression after the simplification. Let us substitute the values of x, y, z in equation (1).
\[\begin{align}
  & {{\left( \sqrt[3]{a} \right)}^{3}}+{{\left( \sqrt[3]{b} \right)}^{3}}+{{\left( \sqrt[3]{c} \right)}^{3}}=3\left( \sqrt[3]{a} \right)\left( \sqrt[3]{b} \right)\left( \sqrt[3]{c} \right) \\
 & a+b+c=3\left( \sqrt[3]{a} \right)\left( \sqrt[3]{b} \right)\left( \sqrt[3]{c} \right) \\
 & a+b+c=3\sqrt[3]{abc}. \\
\end{align}\]
Let’s take cubes on both the sides.
\[\begin{align}
  & {{\left( a+b+c \right)}^{3}}={{\left( 3\sqrt[3]{abc} \right)}^{3}}={{3}^{3}}\left( abc \right) \\
 & {{\left( a+b+c \right)}^{3}}=27abc. \\
\end{align}\]
Thus we got the value of \[{{\left( a+b+c \right)}^{3}}\] as \[27abc.\]
Hence option A is the correct answer.
Note:The key to solving this question is the expansion of \[\left( {{x}^{3}}+{{y}^{3}}+{{z}^{3}} \right)\]. By using this identity, we can easily get a relation and find out the value of \[\left( a+b+c \right)\]. And also remember \[{{\left( \sqrt[3]{a} \right)}^{3}}=a.\]