The term independent of $x$ in expansion of ${\left( {\dfrac{{x + 1}}{{{x^{\dfrac{2}{3}}} - {x^{\dfrac{1}{3}}} + 1}} - \dfrac{{x - 1}}{{x - {x^{\dfrac{1}{2}}}}}} \right)^{10}}$ is:
Answer
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Hint:
Here, we will simplify the given expression by applying different algebraic identities. Then, after simplifying it, we will find the general term of the binomial expansion of the given expression. We will then equate the exponential power of $x$ to 0 to find the term independent of $x$ Then we will substitute the obtained value in the general term of the binomial theorem to find the required term independent of $x$.
Formula Used:
We will use the following formulas:
1) $\left( {{a^3} + {b^3}} \right) = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)$
2) $\left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right)$
3) ${T_{r + 1}} = {}^n{C_r}{a^{n - r}}{b^r}$
4) ${a^m} \times {a^n} = {a^{m + n}}$
5) ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
Complete step by step solution:
Given expression is: ${\left( {\dfrac{{x + 1}}{{{x^{\dfrac{2}{3}}} - {x^{\dfrac{1}{3}}} + 1}} - \dfrac{{x - 1}}{{x - {x^{\dfrac{1}{2}}}}}} \right)^{10}}$
Now, this can also be written as:
${\left( {\dfrac{{x + 1}}{{{x^{\dfrac{2}{3}}} - {x^{\dfrac{1}{3}}} + 1}} - \dfrac{{x - 1}}{{x - {x^{\dfrac{1}{2}}}}}} \right)^{10}} = {\left( {\dfrac{{{{\left( {{x^{\dfrac{1}{3}}}} \right)}^3} + {1^3}}}{{{x^{\dfrac{2}{3}}} - {x^{\dfrac{1}{3}}} + 1}} - \dfrac{{{{\left( {\sqrt x } \right)}^2} - {1^2}}}{{x - {x^{\dfrac{1}{2}}}}}} \right)^{10}}$
Now, taking out $\sqrt x $ common from the denominator of the second fraction, we get,
$ \Rightarrow {\left( {\dfrac{{x + 1}}{{{x^{\dfrac{2}{3}}} - {x^{\dfrac{1}{3}}} + 1}} - \dfrac{{x - 1}}{{x - {x^{\dfrac{1}{2}}}}}} \right)^{10}} = {\left( {\dfrac{{{{\left( {{x^{\dfrac{1}{3}}}} \right)}^3} + {1^3}}}{{{x^{\dfrac{2}{3}}} - {x^{\dfrac{1}{3}}} + 1}} - \dfrac{{{{\left( {\sqrt x } \right)}^2} - {1^2}}}{{\sqrt x \left( {\sqrt x - 1} \right)}}} \right)^{10}}$
Using the properties $\left( {{a^3} + {b^3}} \right) = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)$ and $\left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right)$ in the numerators respectively, we get,
$ \Rightarrow {\left( {\dfrac{{x + 1}}{{{x^{\dfrac{2}{3}}} - {x^{\dfrac{1}{3}}} + 1}} - \dfrac{{x - 1}}{{x - {x^{\dfrac{1}{2}}}}}} \right)^{10}} = {\left( {\dfrac{{\left( {{x^{\dfrac{1}{3}}} + 1} \right)\left( {{x^{\dfrac{2}{3}}} + 1 - {x^{\dfrac{1}{3}}}} \right)}}{{{x^{\dfrac{2}{3}}} - {x^{\dfrac{1}{3}}} + 1}} - \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\sqrt x \left( {\sqrt x - 1} \right)}}} \right)^{10}}$
Cancelling out the same brackets from the numerator and denominator, we get,
$ \Rightarrow {\left( {\dfrac{{x + 1}}{{{x^{\dfrac{2}{3}}} - {x^{\dfrac{1}{3}}} + 1}} - \dfrac{{x - 1}}{{x - {x^{\dfrac{1}{2}}}}}} \right)^{10}} = {\left( {\left( {{x^{\dfrac{1}{3}}} + 1} \right) - \dfrac{{\left( {\sqrt x + 1} \right)}}{{\sqrt x }}} \right)^{10}}$
Splitting the denominator of the second fraction, we get
$ \Rightarrow {\left( {\dfrac{{x + 1}}{{{x^{\dfrac{2}{3}}} - {x^{\dfrac{1}{3}}} + 1}} - \dfrac{{x - 1}}{{x - {x^{\dfrac{1}{2}}}}}} \right)^{10}} = {\left( {{x^{\dfrac{1}{3}}} + 1 - \dfrac{{\sqrt x }}{{\sqrt x }} - \dfrac{1}{{\sqrt x }}} \right)^{10}}$
$ \Rightarrow {\left( {\dfrac{{x + 1}}{{{x^{\dfrac{2}{3}}} - {x^{\dfrac{1}{3}}} + 1}} - \dfrac{{x - 1}}{{x - {x^{\dfrac{1}{2}}}}}} \right)^{10}} = {\left( {{x^{\dfrac{1}{3}}} + 1 - 1 - \dfrac{1}{{{x^{\dfrac{1}{2}}}}}} \right)^{10}}$
Simplifying the expression, we get
$ \Rightarrow {\left( {\dfrac{{x + 1}}{{{x^{\dfrac{2}{3}}} - {x^{\dfrac{1}{3}}} + 1}} - \dfrac{{x - 1}}{{x - {x^{\dfrac{1}{2}}}}}} \right)^{10}} = {\left( {{x^{\dfrac{1}{3}}} - {x^{ - \dfrac{1}{2}}}} \right)^{10}}$
Now we will use the general term of binomial theorem for the given expression ${\left( {\dfrac{{x + 1}}{{{x^{\dfrac{2}{3}}} - {x^{\dfrac{1}{3}}} + 1}} - \dfrac{{x - 1}}{{x - {x^{\dfrac{1}{2}}}}}} \right)^{10}}$.
Here, substituting $n = 10$, $a = {x^{\dfrac{1}{3}}}$ and $b = - {x^{ - \dfrac{1}{2}}}$ in the formula ${T_{r + 1}} = {}^n{C_r}{a^{n - r}}{b^r}$, we get,
${T_{r + 1}} = {}^{10}{C_r}{\left( {{x^{\dfrac{1}{3}}}} \right)^{10 - r}}{\left( { - {x^{ - \dfrac{1}{2}}}} \right)^r}$
$ \Rightarrow {T_{r + 1}} = {}^{10}{C_r}{\left( { - 1} \right)^r}{\left( x \right)^{\dfrac{{10 - r}}{3}}}{\left( x \right)^{ - \dfrac{r}{2}}}$
Now, using the identity ${a^m} \times {a^n} = {a^{m + n}}$, we get,
$ \Rightarrow {T_{r + 1}} = {}^{10}{C_r}{\left( { - 1} \right)^r}{\left( x \right)^{\dfrac{{10 - r}}{3}}}^{ - \dfrac{r}{2}}$……………………………………..$\left( 1 \right)$
Now, for independent of $x$, we will equate the exponent of $x$ to 0.
$\dfrac{{10 - r}}{3} - \dfrac{r}{2} = 0$
Taking LCM and solving further, we get
$ \Rightarrow 20 - 2r - 3r = 0$
Adding and subtracting the like terms, we get
$ \Rightarrow 5r = 20$
Dividing both sides by 5, we get
$ \Rightarrow r = 4$
Hence, substituting $r = 4$ in equation $\left( 1 \right)$, we get,
${T_{4 + 1}} = {}^{10}{C_4}{\left( { - 1} \right)^4}{\left( x \right)^{\dfrac{{10 - 4}}{3}}}^{ - \dfrac{4}{2}}$
Simplifying the expression, we get
$ \Rightarrow {T_5} = {}^{10}{C_4}{\left( x \right)^0} = {}^{10}{C_4}$
Now, using the formula ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, we get,
$ \Rightarrow {T_5} = {}^{10}{C_4} = \dfrac{{10!}}{{4!6!}} = \dfrac{{10 \times 9 \times 8 \times 7}}{{4 \times 3 \times 2}}$
Simplifying the expression further, we get
$ \Rightarrow {T_5} = 210$
Therefore, the term independent of $x$ in expansion of ${\left( {\dfrac{{x + 1}}{{{x^{\dfrac{2}{3}}} - {x^{\dfrac{1}{3}}} + 1}} - \dfrac{{x - 1}}{{x - {x^{\dfrac{1}{2}}}}}} \right)^{10}}$ is 210.
Hence, this is the required answer.
Note:
In algebra, binomial theorem describes the algebraic expansion of the powers of a binomial. Whenever, we are required to find any term independent of $x$ then, we try to make its exponential power 0. This is because if we have ${x^0}$, then by the formula ${x^0} = 1$, our variable $x$ will get vanished. Thus, we equate the power to 0, to find the value of $r$ to be substituted in the general term and hence, find the required term which is independent of $x$.
Here, we will simplify the given expression by applying different algebraic identities. Then, after simplifying it, we will find the general term of the binomial expansion of the given expression. We will then equate the exponential power of $x$ to 0 to find the term independent of $x$ Then we will substitute the obtained value in the general term of the binomial theorem to find the required term independent of $x$.
Formula Used:
We will use the following formulas:
1) $\left( {{a^3} + {b^3}} \right) = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)$
2) $\left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right)$
3) ${T_{r + 1}} = {}^n{C_r}{a^{n - r}}{b^r}$
4) ${a^m} \times {a^n} = {a^{m + n}}$
5) ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
Complete step by step solution:
Given expression is: ${\left( {\dfrac{{x + 1}}{{{x^{\dfrac{2}{3}}} - {x^{\dfrac{1}{3}}} + 1}} - \dfrac{{x - 1}}{{x - {x^{\dfrac{1}{2}}}}}} \right)^{10}}$
Now, this can also be written as:
${\left( {\dfrac{{x + 1}}{{{x^{\dfrac{2}{3}}} - {x^{\dfrac{1}{3}}} + 1}} - \dfrac{{x - 1}}{{x - {x^{\dfrac{1}{2}}}}}} \right)^{10}} = {\left( {\dfrac{{{{\left( {{x^{\dfrac{1}{3}}}} \right)}^3} + {1^3}}}{{{x^{\dfrac{2}{3}}} - {x^{\dfrac{1}{3}}} + 1}} - \dfrac{{{{\left( {\sqrt x } \right)}^2} - {1^2}}}{{x - {x^{\dfrac{1}{2}}}}}} \right)^{10}}$
Now, taking out $\sqrt x $ common from the denominator of the second fraction, we get,
$ \Rightarrow {\left( {\dfrac{{x + 1}}{{{x^{\dfrac{2}{3}}} - {x^{\dfrac{1}{3}}} + 1}} - \dfrac{{x - 1}}{{x - {x^{\dfrac{1}{2}}}}}} \right)^{10}} = {\left( {\dfrac{{{{\left( {{x^{\dfrac{1}{3}}}} \right)}^3} + {1^3}}}{{{x^{\dfrac{2}{3}}} - {x^{\dfrac{1}{3}}} + 1}} - \dfrac{{{{\left( {\sqrt x } \right)}^2} - {1^2}}}{{\sqrt x \left( {\sqrt x - 1} \right)}}} \right)^{10}}$
Using the properties $\left( {{a^3} + {b^3}} \right) = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)$ and $\left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right)$ in the numerators respectively, we get,
$ \Rightarrow {\left( {\dfrac{{x + 1}}{{{x^{\dfrac{2}{3}}} - {x^{\dfrac{1}{3}}} + 1}} - \dfrac{{x - 1}}{{x - {x^{\dfrac{1}{2}}}}}} \right)^{10}} = {\left( {\dfrac{{\left( {{x^{\dfrac{1}{3}}} + 1} \right)\left( {{x^{\dfrac{2}{3}}} + 1 - {x^{\dfrac{1}{3}}}} \right)}}{{{x^{\dfrac{2}{3}}} - {x^{\dfrac{1}{3}}} + 1}} - \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\sqrt x \left( {\sqrt x - 1} \right)}}} \right)^{10}}$
Cancelling out the same brackets from the numerator and denominator, we get,
$ \Rightarrow {\left( {\dfrac{{x + 1}}{{{x^{\dfrac{2}{3}}} - {x^{\dfrac{1}{3}}} + 1}} - \dfrac{{x - 1}}{{x - {x^{\dfrac{1}{2}}}}}} \right)^{10}} = {\left( {\left( {{x^{\dfrac{1}{3}}} + 1} \right) - \dfrac{{\left( {\sqrt x + 1} \right)}}{{\sqrt x }}} \right)^{10}}$
Splitting the denominator of the second fraction, we get
$ \Rightarrow {\left( {\dfrac{{x + 1}}{{{x^{\dfrac{2}{3}}} - {x^{\dfrac{1}{3}}} + 1}} - \dfrac{{x - 1}}{{x - {x^{\dfrac{1}{2}}}}}} \right)^{10}} = {\left( {{x^{\dfrac{1}{3}}} + 1 - \dfrac{{\sqrt x }}{{\sqrt x }} - \dfrac{1}{{\sqrt x }}} \right)^{10}}$
$ \Rightarrow {\left( {\dfrac{{x + 1}}{{{x^{\dfrac{2}{3}}} - {x^{\dfrac{1}{3}}} + 1}} - \dfrac{{x - 1}}{{x - {x^{\dfrac{1}{2}}}}}} \right)^{10}} = {\left( {{x^{\dfrac{1}{3}}} + 1 - 1 - \dfrac{1}{{{x^{\dfrac{1}{2}}}}}} \right)^{10}}$
Simplifying the expression, we get
$ \Rightarrow {\left( {\dfrac{{x + 1}}{{{x^{\dfrac{2}{3}}} - {x^{\dfrac{1}{3}}} + 1}} - \dfrac{{x - 1}}{{x - {x^{\dfrac{1}{2}}}}}} \right)^{10}} = {\left( {{x^{\dfrac{1}{3}}} - {x^{ - \dfrac{1}{2}}}} \right)^{10}}$
Now we will use the general term of binomial theorem for the given expression ${\left( {\dfrac{{x + 1}}{{{x^{\dfrac{2}{3}}} - {x^{\dfrac{1}{3}}} + 1}} - \dfrac{{x - 1}}{{x - {x^{\dfrac{1}{2}}}}}} \right)^{10}}$.
Here, substituting $n = 10$, $a = {x^{\dfrac{1}{3}}}$ and $b = - {x^{ - \dfrac{1}{2}}}$ in the formula ${T_{r + 1}} = {}^n{C_r}{a^{n - r}}{b^r}$, we get,
${T_{r + 1}} = {}^{10}{C_r}{\left( {{x^{\dfrac{1}{3}}}} \right)^{10 - r}}{\left( { - {x^{ - \dfrac{1}{2}}}} \right)^r}$
$ \Rightarrow {T_{r + 1}} = {}^{10}{C_r}{\left( { - 1} \right)^r}{\left( x \right)^{\dfrac{{10 - r}}{3}}}{\left( x \right)^{ - \dfrac{r}{2}}}$
Now, using the identity ${a^m} \times {a^n} = {a^{m + n}}$, we get,
$ \Rightarrow {T_{r + 1}} = {}^{10}{C_r}{\left( { - 1} \right)^r}{\left( x \right)^{\dfrac{{10 - r}}{3}}}^{ - \dfrac{r}{2}}$……………………………………..$\left( 1 \right)$
Now, for independent of $x$, we will equate the exponent of $x$ to 0.
$\dfrac{{10 - r}}{3} - \dfrac{r}{2} = 0$
Taking LCM and solving further, we get
$ \Rightarrow 20 - 2r - 3r = 0$
Adding and subtracting the like terms, we get
$ \Rightarrow 5r = 20$
Dividing both sides by 5, we get
$ \Rightarrow r = 4$
Hence, substituting $r = 4$ in equation $\left( 1 \right)$, we get,
${T_{4 + 1}} = {}^{10}{C_4}{\left( { - 1} \right)^4}{\left( x \right)^{\dfrac{{10 - 4}}{3}}}^{ - \dfrac{4}{2}}$
Simplifying the expression, we get
$ \Rightarrow {T_5} = {}^{10}{C_4}{\left( x \right)^0} = {}^{10}{C_4}$
Now, using the formula ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, we get,
$ \Rightarrow {T_5} = {}^{10}{C_4} = \dfrac{{10!}}{{4!6!}} = \dfrac{{10 \times 9 \times 8 \times 7}}{{4 \times 3 \times 2}}$
Simplifying the expression further, we get
$ \Rightarrow {T_5} = 210$
Therefore, the term independent of $x$ in expansion of ${\left( {\dfrac{{x + 1}}{{{x^{\dfrac{2}{3}}} - {x^{\dfrac{1}{3}}} + 1}} - \dfrac{{x - 1}}{{x - {x^{\dfrac{1}{2}}}}}} \right)^{10}}$ is 210.
Hence, this is the required answer.
Note:
In algebra, binomial theorem describes the algebraic expansion of the powers of a binomial. Whenever, we are required to find any term independent of $x$ then, we try to make its exponential power 0. This is because if we have ${x^0}$, then by the formula ${x^0} = 1$, our variable $x$ will get vanished. Thus, we equate the power to 0, to find the value of $r$ to be substituted in the general term and hence, find the required term which is independent of $x$.
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