Question

Using suitable identity, find the value of:
\[\dfrac{86\times 86\times 86+14\times 14\times 14}{86\times 86-86\times 14+14\times 14}\]

Answer 1

Hint: We can write the given expression as \[\dfrac{{{\left( 86 \right)}^{3}}+{{\left( 14 \right)}^{3}}}{{{\left( 86 \right)}^{2}}-\left( 86 \right)\left( 14 \right)-{{\left( 14 \right)}^{2}}}\].
Use the identity $''{{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)''$to simplify the given expression and find the required value. Take a = 86 and b = 14 for simplifying.

Complete step-by-step answer:
Given expression is: \[\dfrac{86\times 86\times 86+14\times 14\times 14}{86\times 86-86\times 14+14\times 14}\]
We can write the given expression as \[\dfrac{{{\left( 86 \right)}^{3}}+{{\left( 14 \right)}^{3}}}{{{\left( 86 \right)}^{2}}-\left( 86 \right)\left( 14 \right)-{{\left( 14 \right)}^{2}}}\].
As in question, it is clearly mentioned that we have to find the value of a given expression using any suitable identity, so we will not solve it directly but we will first find an identity which can be used for simplifying the given expression.
We know the identity $''{{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)''$.
In this identity let us put a = 86 and b = 14.
After putting a = 86 and b = 14, we will get,
${{\left( 86 \right)}^{3}}+{{\left( 14 \right)}^{3}}=\left( 86+14 \right)\left( {{\left( 86 \right)}^{2}}-\left( 86 \right)\left( 14 \right)+{{\left( 14 \right)}^{2}} \right)..........\left( 1 \right)$
In this identity both the numerator and denominator of given expression is involved. So, this will be a suitable identity. Using this identity we can easily compute the value of a given expression.
On replacing ${{\left( 86 \right)}^{3}}+{{\left( 14 \right)}^{3}}$ in the numerator of given expression with $\left( 86+14 \right)\left( {{\left( 86 \right)}^{2}}-\left( 86 \right)\left( 14 \right)+{{\left( 14 \right)}^{2}} \right)$ from equation (1), we will get,
\[\begin{align}
  & \dfrac{86\times 86\times 86+14\times 14\times 14}{86\times 86-86\times 14+14\times 14}=\dfrac{{{\left( 86 \right)}^{3}}+{{\left( 14 \right)}^{3}}}{{{\left( 86 \right)}^{2}}-\left( 86 \right)\left( 14 \right)-{{\left( 14 \right)}^{2}}} \\
 & =\dfrac{\left( 86+14 \right)\left( {{\left( 86 \right)}^{2}}-\left( 86 \right)\left( 14 \right)+{{\left( 14 \right)}^{2}} \right)}{{{\left( 86 \right)}^{2}}-\left( 86 \right)\left( 14 \right)-{{\left( 14 \right)}^{2}}}\ \ \left[ \text{using equation }\left( 1 \right) \right] \\
\end{align}\]
On dividing both numerator and denominator by $\left[ {{\left( 86 \right)}^{2}}-\left( 86 \right)\left( 14 \right)+{{\left( 14 \right)}^{2}} \right]$, we will get,
\[\begin{align}
  & \dfrac{86\times 86\times 86+14\times 14\times 14}{86\times 86-86\times 14+14\times 14}=\left( 86+14 \right) \\
 & =100 \\
\end{align}\]
So, the required value of \[\dfrac{86\times 86\times 86+14\times 14\times 14}{86\times 86-86\times 14+14\times 14}\] is 100.

Note: In our solution, we have used the identity $''{{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)''$. We can prove this identity in the following way,
$LHS={{a}^{3}}+{{b}^{3}}$
Adding and subtracting 3ab (a + b), we will get,
$LHS={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)-3ab\left( a+b \right)$
We know, ${{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)$
On replacing ${{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)$ with ${{\left( a+b \right)}^{3}}$ in above equation, we will get,
$LHS={{\left( a+b \right)}^{3}}-3ab\left( a+b \right)$
Taking $\left( a+b \right)$ common from both the terms, we will get,
$LHS=\left( a+b \right)\left[ {{\left( a+b \right)}^{2}}-3ab \right]$
We know, ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$
Replacing ${{\left( a+b \right)}^{2}}$ with ${{a}^{2}}+{{b}^{2}}+2ab$ in above equation, we will get,
$\begin{align}
  & LHS=\left( a+b \right)\left[ {{a}^{2}}+{{b}^{2}}+2ab-3ab \right] \\
 & \Rightarrow LHS=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right) \\
 & \Rightarrow LHS=RHS \\
\end{align}$