$${\left( {{{x + 1} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{x - 1} \over {x - {x^{1/2}}}}} \right)^{10}}$$

= $${\left( {{{{{\left( {{x^{1/3}}} \right)}^3} + {{\left( {{1^{1/3}}} \right)}^3}} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{{{\left( {\sqrt x } \right)}^2} - {{\left( 1 \right)}^2}} \over {x - {x^{1/2}}}}} \right)^{10}}$$

= $${\left( {{{\left( {{x^{1/3}} + 1} \right)\left( {{x^{2/3}} - {x^{1/3}} + 1} \right)} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)} \over {\sqrt x \left( {\sqrt x - 1} \right)}}} \right)^{10}}$$

= $${\left( {\left( {{x^{1/3}} + 1} \right) - {{\left( {\sqrt x + 1} \right)} \over {\sqrt x }}} \right)^{10}}$$

= $${\left( {\left( {{x^{1/3}} + 1} \right) - \left( {1 + {1 \over {\sqrt x }}} \right)} \right)^{10}}$$

= $${\left( {{x^{1/3}} - {1 \over {{x^{1/2}}}}} \right)^{10}}$$

[Note:

For $${\left( {{x^\alpha } \pm {1 \over {{x^\beta }}}} \right)^n}$$ the $$\left( {r + 1} \right)$$^th term with power m of x is

$$r = {{n\alpha - m} \over {\alpha + \beta }}$$]

Here $$\alpha = {1 \over 3}$$, $$\beta = {1 \over 2}$$ and m = 0

then $$r = {{10 \times {1 \over 3} - 0} \over {{1 \over 3} + {1 \over 2}}}$$ = $${{10} \over 3} \times {6 \over 5}$$ = 4

$$\therefore$$ T₅ is the term independent of x.

$$\therefore$$ T₅ = $${}^{10}{C_4}$$ = 210

Let $${\left( {a + x} \right)^n}$$ = Odd trems(A) + Even terms(B)

So $${\left( {a - x} \right)^n}$$ = Odd terms(A) - Even terms(B)

$$\therefore$$ $${\left( {a + x} \right)^n} - {\left( {a - x} \right)^n}$$

= (A + B) - (A - B)

= 2B

= 2[even terms]

= 2[ T₂ + T₄ + T₆ + ....... ]

So in case of $${\left( {\sqrt 3 + 1} \right)^{2n}} - {\left( {\sqrt 3 - 1} \right)^{2n}}$$

= 2[ T₂ + T₄ + T₆ + ....... ]

= 2[ $${}^{2n}{C_1}.{\left( {\sqrt 3 } \right)^{2n - 1}} + {}^{2n}{C_3}.{\left( {\sqrt 3 } \right)^{2n - 3}} + ...$$

Here in $${\left( {\sqrt 3 } \right)^{2n - 1}}$$, 2n - 1 is odd number. So there will be always $${\sqrt 3 }$$ in $${\left( {\sqrt 3 } \right)^{2n - 1}}$$.

So $${\left( {\sqrt 3 + 1} \right)^{2n}} - {\left( {\sqrt 3 - 1} \right)^{2n}}$$ will be always irrational number.

Given,
$${\left( {1 - x - {x^2} + {x^3}} \right)^6}$$

= $${\left[ {\left( {1 - x} \right) - {x^2}\left( {1 - x} \right)} \right]^6}$$

= $${\left( {1 - x} \right)^6}{\left( {1 - {x^2}} \right)^6}$$

= $$\left( {1 + {}^6{C_1}( - x) + {}^6{C_2}{{( - x)}^2} + {}^6{C_3}{{( - x)}^3} + .......} \right)\times$$
        $$\left( {1 + {}^6{C_1}( - {x^2}) + {}^6{C_2}{{( - {x^2})}^2} + {}^6{C_3}{{( - {x^2})}^3} + .......} \right)$$

$$\therefore$$ Coefficient of x⁷ = $$ - {}^6{C_1} \times - {}^6{C_3} + \left( { - {}^6{C_3}} \right) \times {}^6{C_2} + \left( { - {}^6{C_5}} \right) \times - {}^6{C_1}$$

= 120 - 300 + 36 = - 144

Note :

$$\sum\limits_{r = 0}^n {r.{}^n{C_r}} $$ = $$ = n{.2^{n - 1}}$$

$$\sum\limits_{r = 0}^n {{r^2}.{}^n{C_r}} = n\left( {n + 1} \right){2^{n - 2}}$$

Given that,

$${s_1} = \sum\limits_{j = 1}^{10} {j\left( {j - 1} \right){}^{10}} {C_j}$$

=$$\sum\limits_{j = 1}^{10} {{j^2}.{}^{10}} {C_j} - \sum\limits_{j = 1}^{10} {j.{}^{10}} {C_j}$$

= 10$$ \times $$11$$ \times $$$${2^{10 - 2}}$$ - 10$$ \times $$$${2^{10 - 1}}$$

= 10$$ \times $$$${2^{8}}$$(11 - 2)

= 10$$ \times $$9$$ \times $$$${2^{8}}$$

= 90$$ \times $$$${2^{8}}$$

$${{s_2} = \sum\limits_{j = 1}^{10} {} } j.{}^{10}{C_j}$$

= 10$$ \times $$$${2^{10-1}}$$

= 10$$ \times $$$${2^{9}}$$

$${{s_3} = \sum\limits_{j = 1}^{10} {{j^2}.{}^{10}{C_j}.} }$$

= 10$$ \times $$11$$ \times $$$${2^{10-2}}$$

= $${{110} \over 2} \times {2^9}$$

= 55 $$ \times $$ $$ {2^9}$$