Question

If the expression $\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}$ is valid, then show that $\dfrac{{{x}^{3}}}{{{a}^{3}}}+\dfrac{{{y}^{3}}}{{{b}^{3}}}+\dfrac{{{z}^{3}}}{{{c}^{3}}}=\dfrac{3xyz}{abc}$.

Answer 1

Hint: To prove LHS = RHS, we will first find the value of LHS by expanding it and by doing some algebraic operation on it. Then we will find RHS and try to make it equal to LHS. We can start solving this question by assuming $\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=k$.

Complete step-by-step answer:
It is given in the question that $\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}$. We have to show that $\dfrac{{{x}^{3}}}{{{a}^{3}}}+\dfrac{{{y}^{3}}}{{{b}^{3}}}+\dfrac{{{z}^{3}}}{{{c}^{3}}}=\dfrac{3xyz}{abc}$.
Let us assume that $\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=k$………………. (1)
So, from equation (1) we get the value of x = ak, y = bk and z = ck respectively. On equating $\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}$ with k we get;
$\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=\dfrac{ak}{a}=\dfrac{bk}{b}=\dfrac{ck}{c}$
Now putting the value of x = ak, y = bk and z = ck in LHS of the given expression to be proved, we get;
$\dfrac{{{\left( ak \right)}^{3}}}{{{\left( a \right)}^{3}}}+\dfrac{{{\left( bk \right)}^{3}}}{{{\left( b \right)}^{3}}}+\dfrac{{{\left( ck \right)}^{3}}}{{{\left( c \right)}^{3}}}$
On expanding LHS further we get;
$\dfrac{{{a}^{3}}{{k}^{3}}}{{{a}^{3}}}+\dfrac{{{b}^{3}}{{k}^{3}}}{{{b}^{3}}}+\dfrac{{{c}^{3}}{{k}^{3}}}{{{c}^{3}}}$
Cancelling ${{a}^{3}}$from numerator and denominator of $\dfrac{{{a}^{3}}{{k}^{3}}}{{{a}^{3}}}$ in LHS we get ${{k}^{3}}$. Similarly, on cancelling ${{b}^{3}}$ from numerator and denominator of $\dfrac{{{b}^{3}}{{k}^{3}}}{{{b}^{3}}}$ in LHS we get ${{k}^{3}}$. Also, on cancelling ${{c}^{3}}$ from numerator and denominator of $\dfrac{{{c}^{3}}{{k}^{3}}}{{{c}^{3}}}$ in LHS we get ${{k}^{3}}$.
Now, we get $\dfrac{{{a}^{3}}{{k}^{3}}}{{{a}^{3}}}={{k}^{3}}$, $\dfrac{{{b}^{3}}{{k}^{3}}}{{{b}^{3}}}={{k}^{3}}$ and $\dfrac{{{c}^{3}}{{k}^{3}}}{{{c}^{3}}}={{k}^{3}}$. So, on putting $\dfrac{{{a}^{3}}{{k}^{3}}}{{{a}^{3}}}+\dfrac{{{b}^{3}}{{k}^{3}}}{{{b}^{3}}}+\dfrac{{{c}^{3}}{{k}^{3}}}{{{c}^{3}}}$ as ${{k}^{3}}+{{k}^{3}}+{{k}^{3}}$ in LHS we get;
$LHS={{k}^{3}}+{{k}^{3}}+{{k}^{3}}$
Also, we can write this as $3{{k}^{3}}$ as adding of ${{k}^{3}}+{{k}^{3}}+{{k}^{3}}$ is $3{{k}^{3}}$.
So, we get, $LHS=3{{k}^{3}}$.
Now, we will solve the RHS of the given expression to be proved. The RHS of the given expression is $\dfrac{3xyz}{abc}$.
So, on putting the value of x = (ak), y = (bk) and z = (ck) in RHS we get,
$\dfrac{3\left( ak \right)\left( bk \right)\left( ck \right)}{abc}$
On cancelling a, b, c from numerator and denominator in RHS, we get,
$\dfrac{3\left( ak \right)\left( bk \right)\left( ck \right)}{abc}=3\left( k \right)\left( k \right)\left( k \right)$
Also, we can write $3\left( k \right)\left( k \right)\left( k \right)$ as $3{{k}^{3}}$. So, the value of RHS is $3{{k}^{3}}$.
Now, we will compare LHS and RHS values.
As, we have calculated that LHS is $3{{k}^{3}}$ and RHS is also $3{{k}^{3}}$.
$\begin{align}
  & \Rightarrow 3{{k}^{3}}=3{{k}^{3}} \\
 & \Rightarrow LHS=RHS \\
\end{align}$
So, from this we can say that $LHS=RHS=3{{k}^{3}}$.
Thus,
$\begin{align}
  & \dfrac{{{x}^{3}}}{{{a}^{3}}}+\dfrac{{{y}^{3}}}{{{b}^{3}}}+\dfrac{{{z}^{3}}}{{{c}^{3}}}=\dfrac{3xyz}{abc}=3{{k}^{3}} \\
 & \Rightarrow LHS=RHS \\
\end{align}$
Hence Proved.


Note: We can also solve this question by using an alternate method. It is given that $\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}$.
Let $\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=k$……….. (1)
On cubing LHS we get,
${{\left( \dfrac{x}{a} \right)}^{3}}+{{\left( \dfrac{y}{b} \right)}^{3}}+{{\left( \dfrac{z}{c} \right)}^{3}}$
As, x = ak, y = bk and z = ck. So on putting the value of x = ak, y = bk and z = ck in LHS, we get,
${{\left( \dfrac{ak}{a} \right)}^{3}}+{{\left( \dfrac{bk}{b} \right)}^{3}}+{{\left( \dfrac{ck}{c} \right)}^{3}}$
Cancelling ${{a}^{3}},{{b}^{3}},{{c}^{3}}$ from numerator and denominator, we get,
$\begin{align}
  & LHS={{k}^{3}}+{{k}^{3}}+{{k}^{3}} \\
 & \Rightarrow LHS=3{{k}^{3}} \\
\end{align}$
We can write $3{{\left( k \right)}^{3}}\ as\ 3k.k.k$.
Now, we know that $k=\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}$. So, on putting the value of $k$ as $\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}$ in LHS, we get,
$\begin{align}
  & 3\left( \dfrac{x}{a} \right)\left( \dfrac{y}{b} \right)\left( \dfrac{z}{c} \right) \\
 & \Rightarrow \dfrac{3xyz}{abc} \\
\end{align}$
Also, we have $\dfrac{3xyz}{abc}$ in RHS.
So, from this we have LHS = RHS.
$\Rightarrow \dfrac{3xyz}{abc}=\dfrac{3xyz}{abc}$
Hence proved.