The term independent of 'x' in the expansion of
$${\left( {{{x + 1} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{x - 1} \over {x - {x^{1/2}}}}} \right)^{10}}$$, where x $\ne$ 0, 1 is equal to ______________.
Answer (integer)
210
Solution
$${\left( {{{x + 1} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{x - 1} \over {x - {x^{1/2}}}}} \right)^{10}}$$
<br><br>= $${\left( {{{{{\left( {{x^{1/3}}} \right)}^3} + {{\left( {{1^{1/3}}} \right)}^3}} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{{{\left( {\sqrt x } \right)}^2} - {{\left( 1 \right)}^2}} \over {x - {x^{1/2}}}}} \right)^{10}}$$
<br><br>= $${\left( {{{\left( {{x^{1/3}} + 1} \right)\left( {{x^{2/3}} - {x^{1/3}} + 1} \right)} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)} \over {\sqrt x \left( {\sqrt x - 1} \right)}}} \right)^{10}}$$
<br><br>= $${\left( {\left( {{x^{1/3}} + 1} \right) - {{\left( {\sqrt x + 1} \right)} \over {\sqrt x }}} \right)^{10}}$$
<br><br>= $${\left( {\left( {{x^{1/3}} + 1} \right) - \left( {1 + {1 \over {\sqrt x }}} \right)} \right)^{10}}$$
<br><br>= ${\left( {{x^{1/3}} - {1 \over {{x^{1/2}}}}} \right)^{10}}$
<br><br>[<b>Note:</b>
<br><br>For ${\left( {{x^\alpha } \pm {1 \over {{x^\beta }}}} \right)^n}$ the $\left( {r + 1} \right)$<sup>th</sup> term with power m of x is
<br><br>$r = {{n\alpha - m} \over {\alpha + \beta }}$]
<br><br>Here $\alpha = {1 \over 3}$, $\beta = {1 \over 2}$ and m = 0
<br><br>then $r = {{10 \times {1 \over 3} - 0} \over {{1 \over 3} + {1 \over 2}}}$ = ${{10} \over 3} \times {6 \over 5}$ = 4
<br><br>$\therefore$ T<sub>5</sub> is the term independent of x.
<br><br>$\therefore$ T<sub>5</sub> = ${}^{10}{C_4}$ = 210
About this question
Subject: Mathematics · Chapter: Binomial Theorem · Topic: Binomial Expansion
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