$${\left( {1 - 2\sqrt x } \right)^{50}}$$

= $${}^{50}{C_0} + {}^{50}{C_1}.\left( { - 2\sqrt x } \right) + {}^{50}{C_2}.{\left( { - 2\sqrt x } \right)^2} + ....$$

Now we need to find out those coefficient where degree of x is integer and you can see at odd terms power of x is integer.

Let $${\left( {1 - 2\sqrt x } \right)^{50}}$$ = Odd(A) - Even(B)

So $${\left( {1 + 2\sqrt x } \right)^{50}}$$ = A + B

$$\therefore$$ 2A = $${\left( {1 + 2\sqrt x } \right)^{50}}$$ + $${\left( {1 - 2\sqrt x } \right)^{50}}$$

$$ \Rightarrow A = {1 \over 2}\left[ {{{\left( {1 + 2\sqrt x } \right)}^{50}} + {{\left( {1 - 2\sqrt x } \right)}^{50}}} \right]$$

Now to find sum of coefficient of A, put x = 1.

$$\therefore$$ Sum of coefficient of A = $${1 \over 2}\left[ {{{\left( {1 + 2} \right)}^{50}} + {{\left( {1 - 2} \right)}^{50}}} \right]$$

= $${1 \over 2}\left[ {{{\left( 3 \right)}^{50}} + 1} \right]$$

$$\left( {1 + ax + b{x^2}} \right){\left( {1 - 2x} \right)^{18}}$$

= $${\left( {1 - 2x} \right)^{18}} + ax{\left( {1 - 2x} \right)^{18}} + b{x^2}{\left( {1 - 2x} \right)^{18}}$$

= $$\left( {1 + ax + b{x^2}} \right)\left[ {{}^{18}{C_0} - {}^{18}{C_1}\left( {2x} \right) + {}^{18}{C_2}{{\left( {2x} \right)}^2} - {}^{18}{C_3}{{\left( {2x} \right)}^3} + ....} \right]$$

Coefficient of x³ is
$${\left( { - 2} \right)^3}.{}^{18}{C_3} + a{\left( { - 2} \right)^2}.{}^{18}{C_2} + b\left( { - 2} \right).{}^{18}{C_1}$$ = 0

$$ \Rightarrow 153a - 9b = 1632$$ ....... (1)

Coefficient of x⁴ is
$${\left( { - 2} \right)^4}.{}^{18}{C_4} + a{\left( { - 2} \right)^3}.{}^{18}{C_3} + b{\left( { - 2} \right)^2}.{}^{18}{C_2}$$ = 0

$$ \Rightarrow 3b - 32a = -240$$ ....... (2)

Solving (1) and (2), we get $$a$$ = 16, b = $${{172} \over 3}$$

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

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

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

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

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

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

[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.