Find the number of rational terms present in the expansion of ${{\left( 1+\sqrt{2}+\sqrt[3]{3} \right)}^{6}}$.
(a) 6
(b) 7
(c) 3
(d) 8
Answer
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Hint:We start solving the problem by assuming $a=1+\sqrt{2}$ and $b=\sqrt[3]{3}$. We then recall the expansion of ${{\left( a+b \right)}^{n}}$ and use it for the given expansion. We then substitute$b=\sqrt[3]{3}$ in the obtained result and take out the terms that were not multiplied by the irrational terms. We then substitute $a=1+\sqrt{2}$ in the terms that were multiplied by the rational terms and expand them using the expansion ${{\left( 1+x \right)}^{n}}=1+{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}+......+{}^{n}{{C}_{r}}{{x}^{r}}+......+{}^{n}{{C}_{n}}{{x}^{n}}$. We then separate the rational terms and count them to get the desired result.
Complete step by step answer:
According to the problem, we need to find the total number of rational terms present in the expansion of ${{\left( 1+\sqrt{2}+\sqrt[3]{3} \right)}^{6}}$.
Let us rewrite the terms the terms present in the given expansion as ${{\left( \left( 1+\sqrt{2} \right)+\sqrt[3]{3} \right)}^{6}}$. Let us assume $a=1+\sqrt{2}$ and $b=\sqrt[3]{3}$.
We know that the binomial expansion of ${{\left( a+b \right)}^{n}}$ is defined as ${}^{n}{{C}_{0}}{{a}^{n}}+{}^{n}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+......+{}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}+......+{}^{n}{{C}_{n}}{{b}^{n}}$.
So, we get ${{\left( a+b \right)}^{6}}={}^{6}{{C}_{0}}{{a}^{6}}+{}^{6}{{C}_{1}}{{a}^{6-1}}{{b}^{1}}+{}^{6}{{C}_{2}}{{a}^{6-2}}{{b}^{2}}+{}^{6}{{C}_{3}}{{a}^{6-3}}{{b}^{3}}+{}^{6}{{C}_{4}}{{a}^{6-4}}{{b}^{4}}+{}^{6}{{C}_{5}}{{a}^{6-5}}{{b}^{1}}+{}^{6}{{C}_{6}}{{a}^{6-6}}{{b}^{6}}$.
We know that ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ and $n!=n\times \left( n-1 \right)\times \left( n-2 \right)\times ......\times 2\times 1$.
\[\Rightarrow {{\left( a+b \right)}^{6}}=\dfrac{6!}{0!6!}{{a}^{6}}+\dfrac{6!}{1!\left( 5 \right)!}{{a}^{6-1}}{{b}^{1}}+\dfrac{6!}{2!4!}{{a}^{6-2}}{{b}^{2}}+\dfrac{6!}{3!3!}{{a}^{6-3}}{{b}^{3}}+\dfrac{6!}{4!2!}{{a}^{6-4}}{{b}^{4}}+\dfrac{6!}{5!1!}{{a}^{6-5}}{{b}^{1}}+\dfrac{6!}{6!0!}{{a}^{6-6}}{{b}^{6}}\].
We know that $0!=1$.
\[\Rightarrow {{\left( a+b \right)}^{6}}=\dfrac{1}{1}{{a}^{6}}+\dfrac{6}{1}{{a}^{5}}{{b}^{1}}+\dfrac{6\times 5}{2\times 1}{{a}^{4}}{{b}^{2}}+\dfrac{6\times 5\times 4}{3\times 2\times 1}{{a}^{3}}{{b}^{3}}+\dfrac{6\times 5}{2\times 1}{{a}^{2}}{{b}^{4}}+\dfrac{6}{1}{{a}^{1}}{{b}^{1}}+\dfrac{1}{1}{{a}^{0}}{{b}^{6}}\].
\[\Rightarrow {{\left( a+b \right)}^{6}}={{a}^{6}}+6{{a}^{5}}{{b}^{1}}+15{{a}^{4}}{{b}^{2}}+20{{a}^{3}}{{b}^{3}}+15{{a}^{2}}{{b}^{4}}+6a{{b}^{5}}+{{b}^{6}}\].
Now, let us substitute $b=\sqrt[3]{3}$.
\[\Rightarrow {{\left( a+b \right)}^{6}}={{a}^{6}}+6{{a}^{5}}{{\left( \sqrt[3]{3} \right)}^{1}}+15{{a}^{4}}{{\left( \sqrt[3]{3} \right)}^{2}}+20{{a}^{3}}{{\left( \sqrt[3]{3} \right)}^{3}}+15{{a}^{2}}{{\left( \sqrt[3]{3} \right)}^{4}}+6a{{\left( \sqrt[3]{3} \right)}^{5}}+{{\left( \sqrt[3]{3} \right)}^{6}}\].
\[\Rightarrow {{\left( a+b \right)}^{6}}={{a}^{6}}+6{{a}^{5}}\left( \sqrt[3]{3} \right)+15{{a}^{4}}\left( \sqrt[3]{9} \right)+20{{a}^{3}}\left( 3 \right)+15{{a}^{2}}\left( 3\sqrt[3]{3} \right)+6a\left( 3\sqrt[3]{9} \right)+9\].
\[\Rightarrow {{\left( a+b \right)}^{6}}={{a}^{6}}+6\sqrt[3]{3}{{a}^{5}}+15\sqrt[3]{9}{{a}^{4}}+60{{a}^{3}}+45\sqrt[3]{3}{{a}^{2}}+18\sqrt[3]{9}a+9\]---(1).
We can see that $\sqrt[3]{3}$ and $\sqrt[3]{9}$ are irrational and even if we multiply them with integer, we get irrational only. Since $\sqrt[3]{3}$, $\sqrt[3]{9}$ have different square root comparing with $\sqrt{2}$, we get irrational when they multiply with each other.
So, we now take the terms that are not multiplying with $\sqrt[3]{3}$ and $\sqrt[3]{9}$.
So, we get rational terms only in ${{a}^{6}}$, $60{{a}^{3}}$ and 9(which is already a rational number) ---(2).
Now, we expand ${{a}^{6}}={{\left( 1+\sqrt{2} \right)}^{6}}$. We know that ${{\left( 1+x \right)}^{n}}=1+{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}+......+{}^{n}{{C}_{r}}{{x}^{r}}+......+{}^{n}{{C}_{n}}{{x}^{n}}$.
So, we get ${{a}^{6}}=1+{}^{6}{{C}_{1}}\left( \sqrt{2} \right)+{}^{6}{{C}_{2}}{{\left( \sqrt{2} \right)}^{2}}+{}^{6}{{C}_{3}}{{\left( \sqrt{2} \right)}^{3}}+{}^{6}{{C}_{4}}{{\left( \sqrt{2} \right)}^{4}}+{}^{6}{{C}_{5}}{{\left( \sqrt{2} \right)}^{5}}+{}^{6}{{C}_{6}}{{\left( \sqrt{2} \right)}^{6}}$.
$\Rightarrow {{a}^{6}}=1+\dfrac{6!}{1!5!}\left( \sqrt{2} \right)+\dfrac{6!}{2!4!}\left( 2 \right)+\dfrac{6!}{3!3!}\left( 2\sqrt{2} \right)+\dfrac{6!}{4!2!}\left( 4 \right)+\dfrac{6!}{5!1!}\left( 4\sqrt{2} \right)+\dfrac{6!}{6!0!}\left( 8 \right)$.
$\Rightarrow {{a}^{6}}=1+\dfrac{6}{1}\left( \sqrt{2} \right)+\dfrac{6\times 5}{2\times 1}\left( 2 \right)+\dfrac{6\times 5\times 4}{3\times 2\times 1}\left( 2\sqrt{2} \right)+\dfrac{6\times 5}{2\times 1}\left( 4 \right)+\dfrac{6}{1}\left( 4\sqrt{2} \right)+\dfrac{1}{1}\left( 8 \right)$.
$\Rightarrow {{a}^{6}}=1+6\sqrt{2}+30+40\sqrt{2}+60+24\sqrt{2}+8$.
We have got the terms 1, 30, 60 and 8 as rational which are 4 ---(3).
Now, let us expand ${{a}^{3}}={{\left( 1+\sqrt{2} \right)}^{3}}$.
So, we get ${{a}^{3}}=1+{}^{3}{{C}_{1}}\left( \sqrt{2} \right)+{}^{3}{{C}_{2}}{{\left( \sqrt{2} \right)}^{2}}+{}^{3}{{C}_{3}}{{\left( \sqrt{2} \right)}^{3}}$.
$\Rightarrow {{a}^{3}}=1+\dfrac{3!}{1!2!}\left( \sqrt{2} \right)+\dfrac{3!}{2!1!}\left( 2 \right)+\dfrac{3!}{3!0!}\left( 2\sqrt{2} \right)$.
\[\Rightarrow {{a}^{3}}=1+\dfrac{3}{1}\left( \sqrt{2} \right)+\dfrac{3}{1}\left( 2 \right)+\dfrac{1}{1}\left( 2\sqrt{2} \right)\].
$\Rightarrow {{a}^{3}}=1+3\sqrt{2}+6+2\sqrt{2}$..
We have got the terms 1 and 6 as rational which are 2 ---(4).
From equations (2), (3), (4) we have got a total of $\left( 1+4+2 \right)=7$ rational terms.
∴ We have got a total 7 rational terms in the expansion of ${{\left( 1+\sqrt{2}+\sqrt[3]{3} \right)}^{6}}$.
The correct option for the given problem is (b).
Note:
We get only one rational if we add all the rational numbers we just obtained. Since they were asked about the number of rational terms present in it, we have taken them separately without adding. We can also take $a=1$ and $b=\sqrt{2}+\sqrt[3]{3}$ for solving this problem in which we just need to expand the powers of ‘b’ only to get the result. Similarly, we can expect problems to find the sum of the rational terms present in this expansion.
Complete step by step answer:
According to the problem, we need to find the total number of rational terms present in the expansion of ${{\left( 1+\sqrt{2}+\sqrt[3]{3} \right)}^{6}}$.
Let us rewrite the terms the terms present in the given expansion as ${{\left( \left( 1+\sqrt{2} \right)+\sqrt[3]{3} \right)}^{6}}$. Let us assume $a=1+\sqrt{2}$ and $b=\sqrt[3]{3}$.
We know that the binomial expansion of ${{\left( a+b \right)}^{n}}$ is defined as ${}^{n}{{C}_{0}}{{a}^{n}}+{}^{n}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+......+{}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}+......+{}^{n}{{C}_{n}}{{b}^{n}}$.
So, we get ${{\left( a+b \right)}^{6}}={}^{6}{{C}_{0}}{{a}^{6}}+{}^{6}{{C}_{1}}{{a}^{6-1}}{{b}^{1}}+{}^{6}{{C}_{2}}{{a}^{6-2}}{{b}^{2}}+{}^{6}{{C}_{3}}{{a}^{6-3}}{{b}^{3}}+{}^{6}{{C}_{4}}{{a}^{6-4}}{{b}^{4}}+{}^{6}{{C}_{5}}{{a}^{6-5}}{{b}^{1}}+{}^{6}{{C}_{6}}{{a}^{6-6}}{{b}^{6}}$.
We know that ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ and $n!=n\times \left( n-1 \right)\times \left( n-2 \right)\times ......\times 2\times 1$.
\[\Rightarrow {{\left( a+b \right)}^{6}}=\dfrac{6!}{0!6!}{{a}^{6}}+\dfrac{6!}{1!\left( 5 \right)!}{{a}^{6-1}}{{b}^{1}}+\dfrac{6!}{2!4!}{{a}^{6-2}}{{b}^{2}}+\dfrac{6!}{3!3!}{{a}^{6-3}}{{b}^{3}}+\dfrac{6!}{4!2!}{{a}^{6-4}}{{b}^{4}}+\dfrac{6!}{5!1!}{{a}^{6-5}}{{b}^{1}}+\dfrac{6!}{6!0!}{{a}^{6-6}}{{b}^{6}}\].
We know that $0!=1$.
\[\Rightarrow {{\left( a+b \right)}^{6}}=\dfrac{1}{1}{{a}^{6}}+\dfrac{6}{1}{{a}^{5}}{{b}^{1}}+\dfrac{6\times 5}{2\times 1}{{a}^{4}}{{b}^{2}}+\dfrac{6\times 5\times 4}{3\times 2\times 1}{{a}^{3}}{{b}^{3}}+\dfrac{6\times 5}{2\times 1}{{a}^{2}}{{b}^{4}}+\dfrac{6}{1}{{a}^{1}}{{b}^{1}}+\dfrac{1}{1}{{a}^{0}}{{b}^{6}}\].
\[\Rightarrow {{\left( a+b \right)}^{6}}={{a}^{6}}+6{{a}^{5}}{{b}^{1}}+15{{a}^{4}}{{b}^{2}}+20{{a}^{3}}{{b}^{3}}+15{{a}^{2}}{{b}^{4}}+6a{{b}^{5}}+{{b}^{6}}\].
Now, let us substitute $b=\sqrt[3]{3}$.
\[\Rightarrow {{\left( a+b \right)}^{6}}={{a}^{6}}+6{{a}^{5}}{{\left( \sqrt[3]{3} \right)}^{1}}+15{{a}^{4}}{{\left( \sqrt[3]{3} \right)}^{2}}+20{{a}^{3}}{{\left( \sqrt[3]{3} \right)}^{3}}+15{{a}^{2}}{{\left( \sqrt[3]{3} \right)}^{4}}+6a{{\left( \sqrt[3]{3} \right)}^{5}}+{{\left( \sqrt[3]{3} \right)}^{6}}\].
\[\Rightarrow {{\left( a+b \right)}^{6}}={{a}^{6}}+6{{a}^{5}}\left( \sqrt[3]{3} \right)+15{{a}^{4}}\left( \sqrt[3]{9} \right)+20{{a}^{3}}\left( 3 \right)+15{{a}^{2}}\left( 3\sqrt[3]{3} \right)+6a\left( 3\sqrt[3]{9} \right)+9\].
\[\Rightarrow {{\left( a+b \right)}^{6}}={{a}^{6}}+6\sqrt[3]{3}{{a}^{5}}+15\sqrt[3]{9}{{a}^{4}}+60{{a}^{3}}+45\sqrt[3]{3}{{a}^{2}}+18\sqrt[3]{9}a+9\]---(1).
We can see that $\sqrt[3]{3}$ and $\sqrt[3]{9}$ are irrational and even if we multiply them with integer, we get irrational only. Since $\sqrt[3]{3}$, $\sqrt[3]{9}$ have different square root comparing with $\sqrt{2}$, we get irrational when they multiply with each other.
So, we now take the terms that are not multiplying with $\sqrt[3]{3}$ and $\sqrt[3]{9}$.
So, we get rational terms only in ${{a}^{6}}$, $60{{a}^{3}}$ and 9(which is already a rational number) ---(2).
Now, we expand ${{a}^{6}}={{\left( 1+\sqrt{2} \right)}^{6}}$. We know that ${{\left( 1+x \right)}^{n}}=1+{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}+......+{}^{n}{{C}_{r}}{{x}^{r}}+......+{}^{n}{{C}_{n}}{{x}^{n}}$.
So, we get ${{a}^{6}}=1+{}^{6}{{C}_{1}}\left( \sqrt{2} \right)+{}^{6}{{C}_{2}}{{\left( \sqrt{2} \right)}^{2}}+{}^{6}{{C}_{3}}{{\left( \sqrt{2} \right)}^{3}}+{}^{6}{{C}_{4}}{{\left( \sqrt{2} \right)}^{4}}+{}^{6}{{C}_{5}}{{\left( \sqrt{2} \right)}^{5}}+{}^{6}{{C}_{6}}{{\left( \sqrt{2} \right)}^{6}}$.
$\Rightarrow {{a}^{6}}=1+\dfrac{6!}{1!5!}\left( \sqrt{2} \right)+\dfrac{6!}{2!4!}\left( 2 \right)+\dfrac{6!}{3!3!}\left( 2\sqrt{2} \right)+\dfrac{6!}{4!2!}\left( 4 \right)+\dfrac{6!}{5!1!}\left( 4\sqrt{2} \right)+\dfrac{6!}{6!0!}\left( 8 \right)$.
$\Rightarrow {{a}^{6}}=1+\dfrac{6}{1}\left( \sqrt{2} \right)+\dfrac{6\times 5}{2\times 1}\left( 2 \right)+\dfrac{6\times 5\times 4}{3\times 2\times 1}\left( 2\sqrt{2} \right)+\dfrac{6\times 5}{2\times 1}\left( 4 \right)+\dfrac{6}{1}\left( 4\sqrt{2} \right)+\dfrac{1}{1}\left( 8 \right)$.
$\Rightarrow {{a}^{6}}=1+6\sqrt{2}+30+40\sqrt{2}+60+24\sqrt{2}+8$.
We have got the terms 1, 30, 60 and 8 as rational which are 4 ---(3).
Now, let us expand ${{a}^{3}}={{\left( 1+\sqrt{2} \right)}^{3}}$.
So, we get ${{a}^{3}}=1+{}^{3}{{C}_{1}}\left( \sqrt{2} \right)+{}^{3}{{C}_{2}}{{\left( \sqrt{2} \right)}^{2}}+{}^{3}{{C}_{3}}{{\left( \sqrt{2} \right)}^{3}}$.
$\Rightarrow {{a}^{3}}=1+\dfrac{3!}{1!2!}\left( \sqrt{2} \right)+\dfrac{3!}{2!1!}\left( 2 \right)+\dfrac{3!}{3!0!}\left( 2\sqrt{2} \right)$.
\[\Rightarrow {{a}^{3}}=1+\dfrac{3}{1}\left( \sqrt{2} \right)+\dfrac{3}{1}\left( 2 \right)+\dfrac{1}{1}\left( 2\sqrt{2} \right)\].
$\Rightarrow {{a}^{3}}=1+3\sqrt{2}+6+2\sqrt{2}$..
We have got the terms 1 and 6 as rational which are 2 ---(4).
From equations (2), (3), (4) we have got a total of $\left( 1+4+2 \right)=7$ rational terms.
∴ We have got a total 7 rational terms in the expansion of ${{\left( 1+\sqrt{2}+\sqrt[3]{3} \right)}^{6}}$.
The correct option for the given problem is (b).
Note:
We get only one rational if we add all the rational numbers we just obtained. Since they were asked about the number of rational terms present in it, we have taken them separately without adding. We can also take $a=1$ and $b=\sqrt{2}+\sqrt[3]{3}$ for solving this problem in which we just need to expand the powers of ‘b’ only to get the result. Similarly, we can expect problems to find the sum of the rational terms present in this expansion.
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