Question

Assuming ‘$x$’ to be so small that ${x^2}$ and higher powers of ‘$x$’ can be neglected, show that $\dfrac{{{{\left( {1 + \dfrac{3}{4}x} \right)}^{ - 4}}{{\left( {16 - 3x} \right)}^{\dfrac{1}{2}}}}}{{{{\left( {8 + x} \right)}^{\dfrac{2}{3}}}}}$ is approximately equal to $1 - \dfrac{{305x}}{{96}}$.

Answer 1

Hint: In the given problem, we have to simplify the expression by neglecting ${x^2}$ and higher power of ‘$x$’ like ${x^3}$, ${x^4} \ldots \ldots $. For this, first we will convert each term in the form of ${\left( {1 + x} \right)^n}$ or ${\left( {1 - x} \right)^n}$. Then, we will use the expansion of ${\left( {1 + x} \right)^n}$ or ${\left( {1 - x} \right)^n}$ whatever is applicable. After neglecting ${x^2}$ and higher power of ‘$x$’, we can say that ${\left( {1 + x} \right)^n}$ is approximately equal to $1 + nx$ and ${\left( {1 - x} \right)^n}$ is approximately equal to $1 - nx$.

Complete step-by-step answer:
In this problem, the given expression is $\dfrac{{{{\left( {1 + \dfrac{3}{4}x} \right)}^{ - 4}}{{\left( {16 - 3x} \right)}^{\dfrac{1}{2}}}}}{{{{\left( {8 + x} \right)}^{\dfrac{2}{3}}}}} \cdots \cdots \left( 1 \right)$. Let us simplify the term ${\left( {16 - 3x} \right)^{\dfrac{1}{2}}}$ by taking $16$ common out. So, we can write
$\Rightarrow {\left( {16 - 3x} \right)^{\dfrac{1}{2}}}$
$ = {\left( {16} \right)^{\dfrac{1}{2}}}{\left( {1 - \dfrac{{3x}}{{16}}} \right)^{\dfrac{1}{2}}}$
$ = 4{\left( {1 - \dfrac{{3x}}{{16}}} \right)^{\dfrac{1}{2}}} \cdots \cdots \left( 2 \right)$
Let us simplify the term ${\left( {8 + x} \right)^{\dfrac{2}{3}}}$ by taking $8$ common out. So, we can write
$\Rightarrow {\left( {8 + x} \right)^{\dfrac{2}{3}}}$
$ = {\left( 8 \right)^{\dfrac{2}{3}}}{\left( {1 + \dfrac{x}{8}} \right)^{\dfrac{2}{3}}}$
$ = {\left( {{2^3}} \right)^{\dfrac{2}{3}}}{\left( {1 + \dfrac{x}{8}} \right)^{\dfrac{2}{3}}}$
$ = 4{\left( {1 + \dfrac{x}{8}} \right)^{\dfrac{2}{3}}} \cdots \cdots \left( 3 \right)$
From $\left( 1 \right)$, $\left( 2 \right)$, $\left( 3 \right)$, we can rewrite the given expression. So, we get
$\Rightarrow \dfrac{{{{\left( {1 + \dfrac{3}{4}x} \right)}^{ - 4}}4{{\left( {1 - \dfrac{{3x}}{{16}}} \right)}^{\dfrac{1}{2}}}}}{{4{{\left( {1 + \dfrac{x}{8}} \right)}^{\dfrac{2}{3}}}}} \cdots \cdots \left( 4 \right)$
Let us simplify the expression $\left( 4 \right)$. So, we get ${\left( {1 + \dfrac{3}{4}x} \right)^{ - \,4}}{\left( {1 - \dfrac{{3x}}{{16}}} \right)^{\dfrac{1}{2}}}{\left( {1 + \dfrac{x}{8}} \right)^{ - \;\dfrac{2}{3}}} \cdots \cdots \left( 5 \right)$
Let us expand each term of expression $\left( 5 \right)$. Note that after neglecting ${x^2}$ and higher powers of ‘$x$’ , we can say that ${\left( {1 + x} \right)^n} \approx 1 + nx$ and ${\left( {1 - x} \right)^n} \approx 1 - nx$. Use this information in $\left( 5 \right)$, so we can write
$\Rightarrow \left[ {1 + \left( { - 4} \right)\left( {\dfrac{{3x}}{4}} \right)} \right]\left[ {1 - \dfrac{1}{2}\left( {\dfrac{{3x}}{{16}}} \right)} \right]\left[ {1 + \left( { - \dfrac{2}{3}} \right)\left( {\dfrac{x}{8}} \right)} \right]$
$ \approx \left( {1 - 3x} \right)\left( {1 - \dfrac{{3x}}{{32}}} \right)\left( {1 - \dfrac{x}{{12}}} \right)$
Let us multiply all terms (brackets) of above expression. Note that we will neglect ${x^2}$ and higher powers of ‘$x$’. So we get
$\left( {1 - \dfrac{{3x}}{{32}} - 3x} \right)\left( {1 - \dfrac{x}{{12}}} \right)$
$ \approx \left( {1 - \dfrac{{99x}}{{32}}} \right)\left( {1 - \dfrac{x}{{12}}} \right)$
$ \approx 1 - \dfrac{x}{{12}} - \dfrac{{99x}}{{32}}$
$ \approx 1 - \dfrac{{8x}}{{96}} - \dfrac{{297x}}{{96}}$
$ \approx 1 - \dfrac{{305}}{{96}}x$

Note: Remember that the expression of ${\left( {1 + x} \right)^n}$ is given by ${\left( {1 + x} \right)^n} = 1 + nx + \dfrac{{n\left( {n - 1} \right)}}{{2!}}{x^2} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{3!}}{x^3} + \cdots \cdots $. Replace $x$ by $ - x$, so we can find expansion of ${\left( {1 - x} \right)^n}$.