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**Hint**: Use the rth term of a binomial expansion formula. For a term to be an integer, its power must be free from square root and cube roots. So expand $ {\left( {\sqrt 3 + \sqrt[3]{2}} \right)^9} $ , and find which term is free from square roots and cube roots.

Binomial expansion of $ {\left( {x + y} \right)^n} $ is $ {}^n{C_0}{x^n}{y^0} + {}^n{C_1}{x^{n - 1}}{y^1} + {}^n{C_2}{x^{n - 2}}{y^2} + ..... + {}^n{C_n}{x^0}{y^n} $

(r+1)th term of a binomial expansion is $ {T_{r + 1}} = {}^n{C_r}{x^{n - r}}{y^r} $

**:**

__Complete step-by-step answer__We are given to find the first integral term in the expansion of $ {\left( {\sqrt 3 + \sqrt[3]{2}} \right)^9} $

On comparing $ {\left( {\sqrt 3 + \sqrt[3]{2}} \right)^9} $ with $ {\left( {x + y} \right)^n} $ , we get the value of x is $ \sqrt 3 = {3^{\dfrac{1}{2}}} $ , the value of y is $ \sqrt[3]{2} = {2^{\dfrac{1}{3}}} $ and the value of n is 9.

Binomial expansion of $ {\left( {x + y} \right)^n} $ is $ {}^n{C_0}{x^n}{y^0} + {}^n{C_1}{x^{n - 1}}{y^1} + {}^n{C_2}{x^{n - 2}}{y^2} + ..... + {}^n{C_n}{x^0}{y^n} $ , where C represents a Combination.

Using binomial expansion, expand the given expression.

$

{\left( {\sqrt 3 + \sqrt[3]{2}} \right)^9} \\

\Rightarrow {\left( {\sqrt 3 + \sqrt[3]{2}} \right)^9} = {}^9{C_0}{\left( {\sqrt 3 } \right)^9}{\left( {\sqrt[3]{2}} \right)^0} + {}^9{C_1}{\left( {\sqrt 3 } \right)^8}{\left( {\sqrt[3]{2}} \right)^1} + {}^9{C_2}{\left( {\sqrt 3 } \right)^7}{\left( {\sqrt[3]{2}} \right)^2} + {}^9{C_3}{\left( {\sqrt 3 } \right)^6}{\left( {\sqrt[3]{2}} \right)^3}.... + {}^9{C_9}{\left( {\sqrt 3 } \right)^0}{\left( {\sqrt[3]{2}} \right)^9} \\

= 1.{\left( {{3^{\dfrac{1}{2}}}} \right)^9}.1 + 9.{\left( {{3^{\dfrac{1}{2}}}} \right)^8}.{\left( {{2^{\dfrac{1}{3}}}} \right)^1} + 36.{\left( {{3^{\dfrac{1}{2}}}} \right)^7}.{\left( {{2^{\dfrac{1}{3}}}} \right)^2} + 84.{\left( {{3^{\dfrac{1}{2}}}} \right)^6}.{\left( {{2^{\dfrac{1}{3}}}} \right)^3} + .... + 1.1.{\left( {{2^{\dfrac{1}{3}}}} \right)^9} \\

= {3^{\dfrac{9}{2}}} + 9.\left( {{3^4}} \right).\left( {{2^{\dfrac{1}{3}}}} \right) + 36.\left( {{3^{\dfrac{7}{2}}}} \right).\left( {{2^{\dfrac{2}{3}}}} \right) + 84.\left( {{3^3}} \right).\left( {{2^1}} \right) + ... + {2^3} \\

$

As we can see in the above expansion, the first term has a fraction power, second term also has a fraction power, third term also has two fraction powers, but the fourth term has integer powers so the fourth term will be an integer.

Fourth term is $ 84 \times {3^3} \times 2 = 4536 $

Therefore, the first integral term in the expansion of $ {\left( {\sqrt 3 + \sqrt[3]{2}} \right)^9} $ is its fourth term.

**So, the correct answer is “Option C”.**

**Note**: Another approach

In a binomial expansion for a term to be an integer, it should have fractional powers; this means the term should not have square roots, cube roots or fourth roots.

So we need to find the first term of the expansion of $ {\left( {\sqrt 3 + \sqrt[3]{2}} \right)^9} $ which is an integer.

(r+1)th term of a binomial expansion $ {\left( {x + y} \right)^n} $ is $ {T_{r + 1}} = {}^n{C_r}{x^{n - r}}{y^r} $

In the same way, (r+1)th term of $ {\left( {\sqrt 3 + \sqrt[3]{2}} \right)^9} $ is

$

{T_{r + 1}} = {}^9{C_r}{\left( {\sqrt 3 } \right)^{9 - r}}{\left( {\sqrt[3]{2}} \right)^r} \\

\Rightarrow {T_{r + 1}} = {}^9{C_r}{\left( {{3^{\dfrac{1}{2}}}} \right)^{9 - r}}{\left( {{2^{\dfrac{1}{3}}}} \right)^r} \\

\Rightarrow {T_{r + 1}} = {}^9{C_r}\left( {{3^{\dfrac{{9 - r}}{2}}}} \right)\left( {{2^{\dfrac{r}{3}}}} \right) \\

$

So here for the term to be integer, $ \dfrac{{9 - r}}{2},\dfrac{r}{3} $ must also be integers.

So for these to be integers, $ \left( {9 - r} \right) $ should be divisible by 2 and r should be divisible by 3

The least number which satisfies this condition is 3, So that 9-3=6 is divisible by 2 and 3 is divisible by 3.

So the first integral term when r is equal to 3 is $ {T_{r + 1}} = {T_{3 + 1}} = {T_4} $ , fourth term.

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