Question

Let x, y, z be positive real numbers such that $x+y+z=12$ and ${{x}^{3}}{{y}^{4}}{{z}^{5}}=\left( 0.1 \right){{\left( 600 \right)}^{3}}$. Then ${{x}^{3}}+{{y}^{3}}+{{z}^{3}}$ is equal to:
A. 258
B. 216
C. 342
D. 270

Answer 1

Hint: We will first try to find the values of x, y, and z individually using the factorisation method. The factorisation method means the splitting of the given numbers such that the multiplication of the new numbers are equal to that of the original number. For example, 10 can be written as $\left( 5\times 2 \right)$. After that we will find the required value with the conditions given in the question.

Complete step-by-step solution -
It is given in the question that $x,y,z$ are positive real numbers. It is also given that, $x+y+z=12$ and ${{x}^{3}}{{y}^{4}}{{z}^{5}}=\left( 0.1 \right){{\left( 600 \right)}^{3}}$, and we have to find the value of ${{x}^{3}}+{{y}^{3}}+{{z}^{3}}$. To solve this question, we will first find out the value of $x,y,z$ individually. We will use the factorisation method for the same. Factorisation is a method in mathematics in which we split the numbers into multiplication factors of smaller numbers such that the multiplication of the new numbers is equal to the original numbers. For example, 9 can be factorised as $\left( 3\times 3 \right)$ and 12 can be factorised as $\left( 2\times 2\times 3 \right),\left( 2\times 6 \right),\left( 4\times 3 \right)$.
Now let us consider the question that has been given to us. We have been given that, $\begin{align}
  & {{x}^{3}}{{y}^{4}}{{z}^{5}}=\left( 0.1 \right){{\left( 600 \right)}^{3}} \\
 & \Rightarrow {{x}^{3}}{{y}^{4}}{{z}^{5}}=\left( 0.1 \right)\left( 600 \right)\left( 600 \right)\left( 600 \right) \\
 & \Rightarrow {{x}^{3}}{{y}^{4}}{{z}^{5}}=60\times 600\times 600 \\
 & \Rightarrow {{x}^{3}}{{y}^{4}}{{z}^{5}}={{6}^{3}}\times {{10}^{5}} \\
\end{align}$
By factorisation of these terms, we get,
${{x}^{3}}{{y}^{4}}{{z}^{5}}={{2}^{3}}\times {{3}^{3}}\times {{2}^{5}}\times {{5}^{5}}$
We know that, ${{a}^{n}}\times {{a}^{m}}$ can be written as $\left( {{a}^{n+m}} \right)$. So, by applying that, we get,
$\begin{align}
  & {{x}^{3}}{{y}^{4}}{{z}^{5}}={{2}^{\left( 3+5 \right)}}\times {{\left( 3 \right)}^{3}}\times {{\left( 5 \right)}^{5}} \\
 & \Rightarrow {{x}^{3}}{{y}^{4}}{{z}^{5}}={{\left( 2 \right)}^{8}}\times {{\left( 3 \right)}^{3}}\times {{\left( 5 \right)}^{5}} \\
 & \Rightarrow {{x}^{3}}{{y}^{4}}{{z}^{5}}={{\left( 4 \right)}^{4}}\times {{\left( 3 \right)}^{3}}\times {{\left( 5 \right)}^{5}} \\
 & \Rightarrow {{x}^{3}}{{y}^{4}}{{z}^{5}}={{\left( 3 \right)}^{3}}\times {{\left( 4 \right)}^{4}}\times {{\left( 5 \right)}^{5}} \\
\end{align}$
So, on comparing the left hand side or LHS and the right hand side or RHS, we get, $x=3,y=4,z=5$. Now we can find the value of ${{x}^{3}}+{{y}^{3}}+{{z}^{3}}$ by substituting them. So, we get,
$\begin{align}
  & {{x}^{3}}+{{y}^{3}}+{{z}^{3}}={{\left( 3 \right)}^{3}}+{{\left( 4 \right)}^{3}}+{{\left( 5 \right)}^{3}} \\
 & \Rightarrow {{x}^{3}}+{{y}^{3}}+{{z}^{3}}=27+64+125 \\
 & {{x}^{3}}+{{y}^{3}}+{{z}^{3}}=216 \\
\end{align}$
Therefore, we get the value of the given expression as 216.
Hence, option B is the correct answer.

Note: We can also find the value by using weighted arithmetic inequality, which is, $\dfrac{{{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}+\ldots \ldots {{m}_{n}}{{x}_{n}}}{{{m}_{1}}+{{m}_{2}}+\ldots \ldots {{m}_{n}}}\ge {{\left( {{x}_{1}}^{m}\times {{x}_{2}}^{{{m}_{2}}}\ldots \ldots {{x}_{n}}^{{{m}_{n}}} \right)}^{\dfrac{1}{{{m}_{1}}+{{m}_{2}}+\ldots \ldots {{m}_{n}}}}}$ . As it involves a lot of calculations, it is not advisable to use the same. Thus, the factorisation method is preferred to solve this question.