Question

Solve the equation: $\dfrac{{{\left( 3x-4 \right)}^{3}}-{{\left( x+1 \right)}^{3}}}{{{\left( 3x-4 \right)}^{3}}+{{\left( x+1 \right)}^{3}}}=\dfrac{61}{189}$

Answer 1

Hint: To simplify the equation given in the question in the simplest possible manner we need to use the method of componendo dividendo, according to which if $\dfrac{a}{b}=\dfrac{c}{d}$ then $\dfrac{a+b}{b-a}=\dfrac{c+d}{d-c}$ . After this, take the cube root of both sides of the equation followed by cross multiplying and solving the equation we get to reach the answer.

Complete step-by-step answer:
The equation given in the question is:
$\dfrac{{{\left( 3x-4 \right)}^{3}}-{{\left( x+1 \right)}^{3}}}{{{\left( 3x-4 \right)}^{3}}+{{\left( x+1 \right)}^{3}}}=\dfrac{61}{189}$
Now, to simplify the equation given in the question, we will use the rule of componendo dividendo, according to which if $\dfrac{a}{b}=\dfrac{c}{d}$ then $\dfrac{a+b}{b-a}=\dfrac{c+d}{d-c}$ .
\[\therefore \dfrac{{{\left( 3x-4 \right)}^{3}}-{{\left( x+1 \right)}^{3}}+{{\left( 3x-4 \right)}^{3}}+{{\left( x+1 \right)}^{3}}}{{{\left( 3x-4 \right)}^{3}}+{{\left( x+1 \right)}^{3}}-\left( {{\left( 3x-4 \right)}^{3}}-{{\left( x+1 \right)}^{3}} \right)}=\dfrac{61+189}{189-61}\]
Simplifying the terms further, we get
\[\Rightarrow \dfrac{{{\left( 3x-4 \right)}^{3}}-{{\left( x+1 \right)}^{3}}+{{\left( 3x-4 \right)}^{3}}+{{\left( x+1 \right)}^{3}}}{{{\left( 3x-4 \right)}^{3}}+{{\left( x+1 \right)}^{3}}-{{\left( 3x-4 \right)}^{3}}+{{\left( x+1 \right)}^{3}}}=\dfrac{250}{128}\]
We can see that there are similar terms which can be cancelled off in the LHS. So, we will get the simplified equation as,
\[\Rightarrow \dfrac{2{{\left( 3x-4 \right)}^{3}}}{2{{\left( x+1 \right)}^{3}}}=\dfrac{125}{64}\]
\[\Rightarrow \dfrac{{{\left( 3x-4 \right)}^{3}}}{{{\left( x+1 \right)}^{3}}}=\dfrac{125}{64}\]
Now, we will take the cube root of both the sides of the equation. On doing so, we get
 \[\sqrt[3]{\dfrac{{{\left( 3x-4 \right)}^{3}}}{{{\left( x+1 \right)}^{3}}}}=\sqrt[3]{\dfrac{125}{64}}\]
\[\Rightarrow \dfrac{\sqrt[3]{{{\left( 3x-4 \right)}^{3}}}}{\sqrt[3]{{{\left( x+1 \right)}^{3}}}}=\dfrac{\sqrt[3]{125}}{\sqrt[3]{64}}\]
Now, we know that the cube root of 125 is 5 and the cube root of 64 is 4. So, if we use this in our equation, we get
 \[\dfrac{3x-4}{x+1}=\dfrac{5}{4}\]
Now we will cross multiply and take all the constant terms to one side of the equation and all the terms with x to the other side. On doing so, we get
\[4\left( 3x-4 \right)=5\left( x+1 \right)\]
\[\Rightarrow 12x-16=5x+5\]
\[\Rightarrow 12x-5x=5+16\]
\[\Rightarrow 7x=21\]
\[\Rightarrow x=3\]
Therefore, the possible value of x is 3.

Note: You could have also solved the question by using the formula ${{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3ab\left( a-b \right)$ and ${{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)$ from the first step itself, but that would be very complicated with lots of terms, making it difficult to solve and increasing the chance of making an error.