Question

Find the number of terms free from radical sign in the expansion of${\left( {1 + \sqrt[3]{3} + \sqrt[7]{7}} \right)^{10}}$.
A.1
B.6
C.11
D.None of the above.

Answer 1

First apply the multinomial theorem of expansion to get the general term of the given expression and then use that general term to find which condition makes any term free from radicals. Then apply the condition and try to approach the desired result.

Complete step-by-step answer:
Consider the given expression:
${\left( {1 + \sqrt[3]{3} + \sqrt[7]{7}} \right)^{10}}$
The goal of the problem is to find the number of terms free from radical sign in the expansion of the given expression.
We will use the multinomial theorem of expansion, then the general term of the expansion is given as:
${\left( {{a_1} + {a_2} + {a_3}... + {a_r}} \right)^n} = \dfrac{{n!}}{{{r_1}!{r_2}!{r_3}...{r_k}!}}a_1^{{r_1}}a_2^{{r_2}}a_3^{{r_3}}...a_k^{{r_k}}$, where${r_1} + {r_2} + {r_3} + ... + {r_k} = n$.
So, the expansion of the expression${\left( {1 + \sqrt[3]{3} + \sqrt[7]{7}} \right)^{10}}$ is given as:
${\left( {1 + \sqrt[3]{3} + \sqrt[7]{7}} \right)^{10}} = \dfrac{{10!}}{{a!b!c!}}{\left( 1 \right)^a}{\left( {\sqrt[3]{3}} \right)^b}{\left( {\sqrt[7]{7}} \right)^c}$, where$a + b + c = 10$.
We know that, we can express$\sqrt[3]{3} = {\left( 3 \right)^{\dfrac{1}{3}}}$and$\sqrt[7]{7} = {\left( 7 \right)^{\dfrac{1}{7}}}$. Then the above expansion can be written as:
${\left( {1 + \sqrt[3]{3} + \sqrt[7]{7}} \right)^{10}} = \dfrac{{10!}}{{a!b!c!}}{\left( 1 \right)^a}{\left[ {{{\left( 3 \right)}^{\dfrac{1}{3}}}} \right]^b}{\left[ {{{\left( 7 \right)}^{\dfrac{1}{7}}}} \right]^c}$
${\left( {1 + \sqrt[3]{3} + \sqrt[7]{7}} \right)^{10}} = \dfrac{{10!}}{{a!b!c!}}{\left( 1 \right)^a}{\left( 3 \right)^{\dfrac{b}{3}}}{\left( 7 \right)^{\dfrac{c}{7}}}$
In order to get the term free from radical, the value of $b$ must be multiple of $3$ and the value of $c$ must be multiple of$7$.
We already have the relation that$a + b + c = 10$ and we also know that the value of $b$ must be multiple of $3$ and the value of $c$ must be multiple of $7$.
The possible values of $b$ are $0,3,6$ and $9$. Then the table below defines the possible combinations.

| b| a| c| Sum(a+b+c)
---|---|---|---|---
1| 0| 10| 0| 10
2| 0| 3| 7| 10
3| 3| 0| 7| 10
4| 3| 7| 0| 10
5| 6| 4| 0| 10
6| 9| 1| 0| 10

So, there are 6 possible combinations that satisfy the conditions.
Therefore, there are $6$ possible terms that are free form radical.
Hence, the option (B) is correct.

Note: We can observe the obtained general term as:
${\left( {1 + \sqrt[3]{3} + \sqrt[7]{7}} \right)^{10}} = \dfrac{{10!}}{{a!b!c!}}{\left( 1 \right)^a}{\left( 3 \right)^{\dfrac{b}{3}}}{\left( 7 \right)^{\dfrac{c}{7}}}$
Take a look on the term ${\left( 3 \right)^{\dfrac{b}{3}}}$, if $b$ is the multiple of $3$ then the denominator of the exponent eliminate the numerator and we get the exponent as a natural number.
Similarly, take a look on the term ${\left( 7 \right)^{\dfrac{c}{7}}}$, if $c$ is the multiple of $7$ then the denominator of the exponent eliminate the numerator and we get the exponent as a natural number.
So, this condition gives the terms that are free from exponent.