Find the degree of the given polynomial ${{\left[ x+\sqrt{{{x}^{3}}-1} \right]}^{6}}+{{\left[ x-\sqrt{{{x}^{3}}-1} \right]}^{6}}$?
Answer
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Hint: We start solving the problem by recalling the binomial expansion of ${{\left( a+b \right)}^{n}}$. We then expand ${{\left[ x+\sqrt{{{x}^{3}}-1} \right]}^{6}}$ and ${{\left[ x-\sqrt{{{x}^{3}}-1} \right]}^{6}}$ using the expansion of ${{\left( a+b \right)}^{n}}$. We then add both the expansions and use the formulas $\sqrt{a}={{a}^{\dfrac{1}{2}}}$, ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ to proceed through the problem. We then need calculations and use the fact that the degree of the polynomial is defined as the highest power of the independent variable present in that polynomial to get the required answer.
Complete step-by-step answer:
According to the problem, we need to find the degree of the given polynomial ${{\left[ x+\sqrt{{{x}^{3}}-1} \right]}^{6}}+{{\left[ x-\sqrt{{{x}^{3}}-1} \right]}^{6}}$.
Let us first expand ${{\left[ x+\sqrt{{{x}^{3}}-1} \right]}^{6}}$ and ${{\left[ x-\sqrt{{{x}^{3}}-1} \right]}^{6}}$ separately and then add them to find the degree of the polynomial.
So, we have the polynomial ${{\left[ x+\sqrt{{{x}^{3}}-1} \right]}^{6}}$. Let us assume $a=x$ and $b=\sqrt{{{x}^{3}}-1}$. We know that the binomial expansion of ${{\left( a+b \right)}^{n}}$ is ${}^{n}{{C}_{0}}{{a}^{n}}+{}^{n}{{C}_{1}}{{a}^{n-1}}b+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+......+{}^{n}{{C}_{n-1}}{{a}^{1}}{{b}^{n-1}}+{}^{n}{{C}_{n}}{{b}^{n}}$. Let us use this to expand ${{\left[ x+\sqrt{{{x}^{3}}-1} \right]}^{6}}={{\left[ a+b \right]}^{6}}$.
So, we have \[{{\left[ a+b \right]}^{6}}={}^{6}{{C}_{0}}{{a}^{6}}+{}^{6}{{C}_{1}}{{a}^{5}}b+{}^{6}{{C}_{2}}{{a}^{4}}{{b}^{2}}+{}^{6}{{C}_{3}}{{a}^{3}}{{b}^{3}}+{}^{6}{{C}_{4}}{{a}^{2}}{{b}^{4}}+{}^{6}{{C}_{5}}a{{b}^{5}}+{}^{6}{{C}_{6}}{{b}^{6}}\] ---(1).
Now, we have the polynomial ${{\left[ x-\sqrt{{{x}^{3}}-1} \right]}^{6}}$. Let us assume $a=x$ and $b=\sqrt{{{x}^{3}}-1}$. We know that the binomial expansion of ${{\left( a-b \right)}^{n}}$ is ${}^{n}{{C}_{0}}{{a}^{n}}-{}^{n}{{C}_{1}}{{a}^{n-1}}b+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}-......+{{\left( -1 \right)}^{n-1}}{}^{n}{{C}_{n-1}}{{a}^{1}}{{b}^{n-1}}+{{\left( -1 \right)}^{n}}{}^{n}{{C}_{n}}{{b}^{n}}$. Let us use this to expand ${{\left[ x-\sqrt{{{x}^{3}}-1} \right]}^{6}}={{\left[ a-b \right]}^{6}}$.
So, we have \[{{\left[ a-b \right]}^{6}}={}^{6}{{C}_{0}}{{a}^{6}}-{}^{6}{{C}_{1}}{{a}^{5}}b+{}^{6}{{C}_{2}}{{a}^{4}}{{b}^{2}}-{}^{6}{{C}_{3}}{{a}^{3}}{{b}^{3}}+{}^{6}{{C}_{4}}{{a}^{2}}{{b}^{4}}-{}^{6}{{C}_{5}}a{{b}^{5}}+{}^{6}{{C}_{6}}{{b}^{6}}\] ---(2).
Let us add equation (1) and (2).
$\Rightarrow {{\left[ a+b \right]}^{6}}+{{\left[ a-b \right]}^{6}}=2\left( {}^{6}{{C}_{0}}{{a}^{6}}+{}^{6}{{C}_{2}}{{a}^{4}}{{b}^{2}}+{}^{6}{{C}_{4}}{{a}^{2}}{{b}^{4}}+{}^{6}{{C}_{6}}{{b}^{6}} \right)$ ---(3).
Now let us substitute the $a=x$ and $b=\sqrt{{{x}^{3}}-1}$ in equation (3).
So, we get $\Rightarrow {{\left[ x+\sqrt{{{x}^{3}}-1} \right]}^{6}}+{{\left[ x-\sqrt{{{x}^{3}}-1} \right]}^{6}}=2\left( {}^{6}{{C}_{0}}{{x}^{6}}+{}^{6}{{C}_{2}}{{x}^{4}}{{\left( \sqrt{{{x}^{3}}-1} \right)}^{2}}+{}^{6}{{C}_{4}}{{x}^{2}}{{\left( \sqrt{{{x}^{3}}-1} \right)}^{4}}+{}^{6}{{C}_{6}}{{\left( \sqrt{{{x}^{3}}-1} \right)}^{6}} \right)$.
We know that $\sqrt{a}={{a}^{\dfrac{1}{2}}}$.
$\Rightarrow {{\left[ x+\sqrt{{{x}^{3}}-1} \right]}^{6}}+{{\left[ x-\sqrt{{{x}^{3}}-1} \right]}^{6}}=2\left( {}^{6}{{C}_{0}}{{x}^{6}}+{}^{6}{{C}_{2}}{{x}^{4}}{{\left( {{\left( {{x}^{3}}-1 \right)}^{\dfrac{1}{2}}} \right)}^{2}}+{}^{6}{{C}_{4}}{{x}^{2}}{{\left( {{\left( {{x}^{3}}-1 \right)}^{\dfrac{1}{2}}} \right)}^{4}}+{}^{6}{{C}_{6}}{{\left( {{\left( {{x}^{3}}-1 \right)}^{\dfrac{1}{2}}} \right)}^{6}} \right)$.
We know that ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$.
\[\Rightarrow {{\left[ x+\sqrt{{{x}^{3}}-1} \right]}^{6}}+{{\left[ x-\sqrt{{{x}^{3}}-1} \right]}^{6}}=2\left( {}^{6}{{C}_{0}}{{x}^{6}}+{}^{6}{{C}_{2}}{{x}^{4}}\left( {{x}^{3}}-1 \right)+{}^{6}{{C}_{4}}{{x}^{2}}{{\left( {{x}^{3}}-1 \right)}^{2}}+{}^{6}{{C}_{6}}{{\left( {{x}^{3}}-1 \right)}^{3}} \right)\].
\[\Rightarrow {{\left[ x+\sqrt{{{x}^{3}}-1} \right]}^{6}}+{{\left[ x-\sqrt{{{x}^{3}}-1} \right]}^{6}}=2\left( {}^{6}{{C}_{2}}{{x}^{7}}+{}^{6}{{C}_{0}}{{x}^{6}}-{}^{6}{{C}_{2}}{{x}^{4}}+{}^{6}{{C}_{4}}{{x}^{2}}{{\left( {{x}^{3}}-1 \right)}^{2}}+{}^{6}{{C}_{6}}{{\left( {{x}^{3}}-1 \right)}^{3}} \right)\].
We know that \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\].
\[\Rightarrow {{\left[ x+\sqrt{{{x}^{3}}-1} \right]}^{6}}+{{\left[ x-\sqrt{{{x}^{3}}-1} \right]}^{6}}=2\left( {}^{6}{{C}_{2}}{{x}^{7}}+{}^{6}{{C}_{0}}{{x}^{6}}-{}^{6}{{C}_{2}}{{x}^{4}}+{}^{6}{{C}_{4}}{{x}^{2}}\left( {{x}^{6}}-2{{x}^{3}}+1 \right)+{}^{6}{{C}_{6}}{{\left( {{x}^{3}}-1 \right)}^{3}} \right)\].
\[\Rightarrow {{\left[ x+\sqrt{{{x}^{3}}-1} \right]}^{6}}+{{\left[ x-\sqrt{{{x}^{3}}-1} \right]}^{6}}=2\left( {}^{6}{{C}_{0}}{{x}^{6}}+{}^{6}{{C}_{2}}{{x}^{7}}-{}^{6}{{C}_{2}}{{x}^{4}}+{}^{6}{{C}_{4}}{{x}^{8}}-{}^{6}{{C}_{4}}12{{x}^{5}}+{}^{6}{{C}_{4}}{{x}^{2}}+{}^{6}{{C}_{6}}{{\left( {{x}^{3}}-1 \right)}^{3}} \right)\].
\[\Rightarrow {{\left[ x+\sqrt{{{x}^{3}}-1} \right]}^{6}}+{{\left[ x-\sqrt{{{x}^{3}}-1} \right]}^{6}}=2\left( {}^{6}{{C}_{4}}{{x}^{8}}+{}^{6}{{C}_{2}}{{x}^{7}}+{}^{6}{{C}_{0}}{{x}^{6}}-{}^{6}{{C}_{4}}12{{x}^{5}}-{}^{6}{{C}_{2}}{{x}^{4}}+{}^{6}{{C}_{4}}{{x}^{2}}+{}^{6}{{C}_{6}}{{\left( {{x}^{3}}-1 \right)}^{3}} \right)\].
We know that ${{\left( a-b \right)}^{3}}={{a}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}-{{b}^{3}}$.
\[\Rightarrow {{\left[ x+\sqrt{{{x}^{3}}-1} \right]}^{6}}+{{\left[ x-\sqrt{{{x}^{3}}-1} \right]}^{6}}=2\left( {}^{6}{{C}_{4}}{{x}^{8}}+{}^{6}{{C}_{2}}{{x}^{7}}+{}^{6}{{C}_{0}}{{x}^{6}}-{}^{6}{{C}_{4}}12{{x}^{5}}-{}^{6}{{C}_{2}}{{x}^{4}}+{}^{6}{{C}_{4}}{{x}^{2}}+{}^{6}{{C}_{6}}\left( {{x}^{9}}-3{{x}^{6}}+3{{x}^{3}}-1 \right) \right)\].
\[\Rightarrow {{\left[ x+\sqrt{{{x}^{3}}-1} \right]}^{6}}+{{\left[ x-\sqrt{{{x}^{3}}-1} \right]}^{6}}=2\left( {}^{6}{{C}_{4}}{{x}^{8}}+{}^{6}{{C}_{2}}{{x}^{7}}+{}^{6}{{C}_{0}}{{x}^{6}}-{}^{6}{{C}_{4}}12{{x}^{5}}-{}^{6}{{C}_{2}}{{x}^{4}}+{}^{6}{{C}_{4}}{{x}^{2}}+{}^{6}{{C}_{6}}{{x}^{9}}-3{}^{6}{{C}_{6}}{{x}^{6}}+3{}^{6}{{C}_{6}}{{x}^{3}}-{}^{6}{{C}_{6}} \right)\].
\[\Rightarrow {{\left[ x+\sqrt{{{x}^{3}}-1} \right]}^{6}}+{{\left[ x-\sqrt{{{x}^{3}}-1} \right]}^{6}}=2\left( {}^{6}{{C}_{6}}{{x}^{9}}+{}^{6}{{C}_{4}}{{x}^{8}}+{}^{6}{{C}_{2}}{{x}^{7}}+\left( {}^{6}{{C}_{0}}-3{}^{6}{{C}_{6}} \right){{x}^{6}}-{}^{6}{{C}_{4}}12{{x}^{5}}-{}^{6}{{C}_{2}}{{x}^{4}}+3{}^{6}{{C}_{6}}{{x}^{3}}+{}^{6}{{C}_{4}}{{x}^{2}}-{}^{6}{{C}_{6}} \right)\] ---(4).
We know that the degree of the polynomial is defined as the highest power of independent variables present in that polynomial.
In equation (5), we can see that ‘x’ is the independent variable and the highest power of x is 9. This gives us the degree of the polynomial as 9.
∴ The degree of the polynomial ${{\left[ x+\sqrt{{{x}^{3}}-1} \right]}^{6}}+{{\left[ x-\sqrt{{{x}^{3}}-1} \right]}^{6}}$ is 9.
Note: We can also solve this problem by taking ${{x}^{\dfrac{3}{2}}}$ common from both the terms as shown below.
We have \[{{\left[ x+\sqrt{{{x}^{3}}-1} \right]}^{6}}+{{\left[ x-\sqrt{{{x}^{3}}-1} \right]}^{6}}={{\left[ x+\sqrt{{{x}^{3}}\left( 1-\dfrac{1}{{{x}^{3}}} \right)} \right]}^{6}}+{{\left[ x-\sqrt{{{x}^{3}}\left( 1-\dfrac{1}{{{x}^{3}}} \right)} \right]}^{6}}\].
We know that $\sqrt{{{a}^{m}}b}={{a}^{\dfrac{m}{2}}}\sqrt{b}$.
\[\Rightarrow {{\left[ x+\sqrt{{{x}^{3}}-1} \right]}^{6}}+{{\left[ x-\sqrt{{{x}^{3}}-1} \right]}^{6}}={{\left[ x+{{x}^{\dfrac{3}{2}}}\sqrt{\left( 1-\dfrac{1}{{{x}^{3}}} \right)} \right]}^{6}}+{{\left[ x-{{x}^{\dfrac{3}{2}}}\sqrt{\left( 1-\dfrac{1}{{{x}^{3}}} \right)} \right]}^{6}}\].
\[\Rightarrow {{\left[ x+\sqrt{{{x}^{3}}-1} \right]}^{6}}+{{\left[ x-\sqrt{{{x}^{3}}-1} \right]}^{6}}={{\left[ {{x}^{\dfrac{3}{2}}}\left( \dfrac{1}{{{x}^{\dfrac{1}{2}}}}+\sqrt{\left( 1-\dfrac{1}{{{x}^{3}}} \right)} \right) \right]}^{6}}+{{\left[ {{x}^{\dfrac{3}{2}}}\left( \dfrac{1}{{{x}^{\dfrac{1}{2}}}}-\sqrt{\left( 1-\dfrac{1}{{{x}^{3}}} \right)} \right) \right]}^{6}}\].
We know that ${{\left( ab \right)}^{m}}={{a}^{m}}{{b}^{m}}$.
\[\Rightarrow {{\left[ x+\sqrt{{{x}^{3}}-1} \right]}^{6}}+{{\left[ x-\sqrt{{{x}^{3}}-1} \right]}^{6}}={{\left( {{x}^{\dfrac{3}{2}}} \right)}^{6}}{{\left( \dfrac{1}{{{x}^{\dfrac{1}{2}}}}+\sqrt{\left( 1-\dfrac{1}{{{x}^{3}}} \right)} \right)}^{6}}+{{\left( {{x}^{\dfrac{3}{2}}} \right)}^{6}}{{\left( \dfrac{1}{{{x}^{\dfrac{1}{2}}}}-\sqrt{\left( 1-\dfrac{1}{{{x}^{3}}} \right)} \right)}^{6}}\].
\[\Rightarrow {{\left[ x+\sqrt{{{x}^{3}}-1} \right]}^{6}}+{{\left[ x-\sqrt{{{x}^{3}}-1} \right]}^{6}}=\left( {{x}^{9}} \right){{\left( \dfrac{1}{{{x}^{\dfrac{1}{2}}}}+\sqrt{\left( 1-\dfrac{1}{{{x}^{3}}} \right)} \right)}^{6}}+\left( {{x}^{9}} \right){{\left( \dfrac{1}{{{x}^{\dfrac{1}{2}}}}-\sqrt{\left( 1-\dfrac{1}{{{x}^{3}}} \right)} \right)}^{6}}\].
\[\Rightarrow {{\left[ x+\sqrt{{{x}^{3}}-1} \right]}^{6}}+{{\left[ x-\sqrt{{{x}^{3}}-1} \right]}^{6}}=\left( {{x}^{9}} \right)\left( {{\left( \dfrac{1}{{{x}^{\dfrac{1}{2}}}}+\sqrt{\left( 1-\dfrac{1}{{{x}^{3}}} \right)} \right)}^{6}}+{{\left( \dfrac{1}{{{x}^{\dfrac{1}{2}}}}-\sqrt{\left( 1-\dfrac{1}{{{x}^{3}}} \right)} \right)}^{6}} \right)\].
We can see that the power of x we get in the expansion will be less than or equal to 0, which makes 9 as the highest power of x as well as the degree of the polynomial.
Complete step-by-step answer:
According to the problem, we need to find the degree of the given polynomial ${{\left[ x+\sqrt{{{x}^{3}}-1} \right]}^{6}}+{{\left[ x-\sqrt{{{x}^{3}}-1} \right]}^{6}}$.
Let us first expand ${{\left[ x+\sqrt{{{x}^{3}}-1} \right]}^{6}}$ and ${{\left[ x-\sqrt{{{x}^{3}}-1} \right]}^{6}}$ separately and then add them to find the degree of the polynomial.
So, we have the polynomial ${{\left[ x+\sqrt{{{x}^{3}}-1} \right]}^{6}}$. Let us assume $a=x$ and $b=\sqrt{{{x}^{3}}-1}$. We know that the binomial expansion of ${{\left( a+b \right)}^{n}}$ is ${}^{n}{{C}_{0}}{{a}^{n}}+{}^{n}{{C}_{1}}{{a}^{n-1}}b+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+......+{}^{n}{{C}_{n-1}}{{a}^{1}}{{b}^{n-1}}+{}^{n}{{C}_{n}}{{b}^{n}}$. Let us use this to expand ${{\left[ x+\sqrt{{{x}^{3}}-1} \right]}^{6}}={{\left[ a+b \right]}^{6}}$.
So, we have \[{{\left[ a+b \right]}^{6}}={}^{6}{{C}_{0}}{{a}^{6}}+{}^{6}{{C}_{1}}{{a}^{5}}b+{}^{6}{{C}_{2}}{{a}^{4}}{{b}^{2}}+{}^{6}{{C}_{3}}{{a}^{3}}{{b}^{3}}+{}^{6}{{C}_{4}}{{a}^{2}}{{b}^{4}}+{}^{6}{{C}_{5}}a{{b}^{5}}+{}^{6}{{C}_{6}}{{b}^{6}}\] ---(1).
Now, we have the polynomial ${{\left[ x-\sqrt{{{x}^{3}}-1} \right]}^{6}}$. Let us assume $a=x$ and $b=\sqrt{{{x}^{3}}-1}$. We know that the binomial expansion of ${{\left( a-b \right)}^{n}}$ is ${}^{n}{{C}_{0}}{{a}^{n}}-{}^{n}{{C}_{1}}{{a}^{n-1}}b+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}-......+{{\left( -1 \right)}^{n-1}}{}^{n}{{C}_{n-1}}{{a}^{1}}{{b}^{n-1}}+{{\left( -1 \right)}^{n}}{}^{n}{{C}_{n}}{{b}^{n}}$. Let us use this to expand ${{\left[ x-\sqrt{{{x}^{3}}-1} \right]}^{6}}={{\left[ a-b \right]}^{6}}$.
So, we have \[{{\left[ a-b \right]}^{6}}={}^{6}{{C}_{0}}{{a}^{6}}-{}^{6}{{C}_{1}}{{a}^{5}}b+{}^{6}{{C}_{2}}{{a}^{4}}{{b}^{2}}-{}^{6}{{C}_{3}}{{a}^{3}}{{b}^{3}}+{}^{6}{{C}_{4}}{{a}^{2}}{{b}^{4}}-{}^{6}{{C}_{5}}a{{b}^{5}}+{}^{6}{{C}_{6}}{{b}^{6}}\] ---(2).
Let us add equation (1) and (2).
$\Rightarrow {{\left[ a+b \right]}^{6}}+{{\left[ a-b \right]}^{6}}=2\left( {}^{6}{{C}_{0}}{{a}^{6}}+{}^{6}{{C}_{2}}{{a}^{4}}{{b}^{2}}+{}^{6}{{C}_{4}}{{a}^{2}}{{b}^{4}}+{}^{6}{{C}_{6}}{{b}^{6}} \right)$ ---(3).
Now let us substitute the $a=x$ and $b=\sqrt{{{x}^{3}}-1}$ in equation (3).
So, we get $\Rightarrow {{\left[ x+\sqrt{{{x}^{3}}-1} \right]}^{6}}+{{\left[ x-\sqrt{{{x}^{3}}-1} \right]}^{6}}=2\left( {}^{6}{{C}_{0}}{{x}^{6}}+{}^{6}{{C}_{2}}{{x}^{4}}{{\left( \sqrt{{{x}^{3}}-1} \right)}^{2}}+{}^{6}{{C}_{4}}{{x}^{2}}{{\left( \sqrt{{{x}^{3}}-1} \right)}^{4}}+{}^{6}{{C}_{6}}{{\left( \sqrt{{{x}^{3}}-1} \right)}^{6}} \right)$.
We know that $\sqrt{a}={{a}^{\dfrac{1}{2}}}$.
$\Rightarrow {{\left[ x+\sqrt{{{x}^{3}}-1} \right]}^{6}}+{{\left[ x-\sqrt{{{x}^{3}}-1} \right]}^{6}}=2\left( {}^{6}{{C}_{0}}{{x}^{6}}+{}^{6}{{C}_{2}}{{x}^{4}}{{\left( {{\left( {{x}^{3}}-1 \right)}^{\dfrac{1}{2}}} \right)}^{2}}+{}^{6}{{C}_{4}}{{x}^{2}}{{\left( {{\left( {{x}^{3}}-1 \right)}^{\dfrac{1}{2}}} \right)}^{4}}+{}^{6}{{C}_{6}}{{\left( {{\left( {{x}^{3}}-1 \right)}^{\dfrac{1}{2}}} \right)}^{6}} \right)$.
We know that ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$.
\[\Rightarrow {{\left[ x+\sqrt{{{x}^{3}}-1} \right]}^{6}}+{{\left[ x-\sqrt{{{x}^{3}}-1} \right]}^{6}}=2\left( {}^{6}{{C}_{0}}{{x}^{6}}+{}^{6}{{C}_{2}}{{x}^{4}}\left( {{x}^{3}}-1 \right)+{}^{6}{{C}_{4}}{{x}^{2}}{{\left( {{x}^{3}}-1 \right)}^{2}}+{}^{6}{{C}_{6}}{{\left( {{x}^{3}}-1 \right)}^{3}} \right)\].
\[\Rightarrow {{\left[ x+\sqrt{{{x}^{3}}-1} \right]}^{6}}+{{\left[ x-\sqrt{{{x}^{3}}-1} \right]}^{6}}=2\left( {}^{6}{{C}_{2}}{{x}^{7}}+{}^{6}{{C}_{0}}{{x}^{6}}-{}^{6}{{C}_{2}}{{x}^{4}}+{}^{6}{{C}_{4}}{{x}^{2}}{{\left( {{x}^{3}}-1 \right)}^{2}}+{}^{6}{{C}_{6}}{{\left( {{x}^{3}}-1 \right)}^{3}} \right)\].
We know that \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\].
\[\Rightarrow {{\left[ x+\sqrt{{{x}^{3}}-1} \right]}^{6}}+{{\left[ x-\sqrt{{{x}^{3}}-1} \right]}^{6}}=2\left( {}^{6}{{C}_{2}}{{x}^{7}}+{}^{6}{{C}_{0}}{{x}^{6}}-{}^{6}{{C}_{2}}{{x}^{4}}+{}^{6}{{C}_{4}}{{x}^{2}}\left( {{x}^{6}}-2{{x}^{3}}+1 \right)+{}^{6}{{C}_{6}}{{\left( {{x}^{3}}-1 \right)}^{3}} \right)\].
\[\Rightarrow {{\left[ x+\sqrt{{{x}^{3}}-1} \right]}^{6}}+{{\left[ x-\sqrt{{{x}^{3}}-1} \right]}^{6}}=2\left( {}^{6}{{C}_{0}}{{x}^{6}}+{}^{6}{{C}_{2}}{{x}^{7}}-{}^{6}{{C}_{2}}{{x}^{4}}+{}^{6}{{C}_{4}}{{x}^{8}}-{}^{6}{{C}_{4}}12{{x}^{5}}+{}^{6}{{C}_{4}}{{x}^{2}}+{}^{6}{{C}_{6}}{{\left( {{x}^{3}}-1 \right)}^{3}} \right)\].
\[\Rightarrow {{\left[ x+\sqrt{{{x}^{3}}-1} \right]}^{6}}+{{\left[ x-\sqrt{{{x}^{3}}-1} \right]}^{6}}=2\left( {}^{6}{{C}_{4}}{{x}^{8}}+{}^{6}{{C}_{2}}{{x}^{7}}+{}^{6}{{C}_{0}}{{x}^{6}}-{}^{6}{{C}_{4}}12{{x}^{5}}-{}^{6}{{C}_{2}}{{x}^{4}}+{}^{6}{{C}_{4}}{{x}^{2}}+{}^{6}{{C}_{6}}{{\left( {{x}^{3}}-1 \right)}^{3}} \right)\].
We know that ${{\left( a-b \right)}^{3}}={{a}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}-{{b}^{3}}$.
\[\Rightarrow {{\left[ x+\sqrt{{{x}^{3}}-1} \right]}^{6}}+{{\left[ x-\sqrt{{{x}^{3}}-1} \right]}^{6}}=2\left( {}^{6}{{C}_{4}}{{x}^{8}}+{}^{6}{{C}_{2}}{{x}^{7}}+{}^{6}{{C}_{0}}{{x}^{6}}-{}^{6}{{C}_{4}}12{{x}^{5}}-{}^{6}{{C}_{2}}{{x}^{4}}+{}^{6}{{C}_{4}}{{x}^{2}}+{}^{6}{{C}_{6}}\left( {{x}^{9}}-3{{x}^{6}}+3{{x}^{3}}-1 \right) \right)\].
\[\Rightarrow {{\left[ x+\sqrt{{{x}^{3}}-1} \right]}^{6}}+{{\left[ x-\sqrt{{{x}^{3}}-1} \right]}^{6}}=2\left( {}^{6}{{C}_{4}}{{x}^{8}}+{}^{6}{{C}_{2}}{{x}^{7}}+{}^{6}{{C}_{0}}{{x}^{6}}-{}^{6}{{C}_{4}}12{{x}^{5}}-{}^{6}{{C}_{2}}{{x}^{4}}+{}^{6}{{C}_{4}}{{x}^{2}}+{}^{6}{{C}_{6}}{{x}^{9}}-3{}^{6}{{C}_{6}}{{x}^{6}}+3{}^{6}{{C}_{6}}{{x}^{3}}-{}^{6}{{C}_{6}} \right)\].
\[\Rightarrow {{\left[ x+\sqrt{{{x}^{3}}-1} \right]}^{6}}+{{\left[ x-\sqrt{{{x}^{3}}-1} \right]}^{6}}=2\left( {}^{6}{{C}_{6}}{{x}^{9}}+{}^{6}{{C}_{4}}{{x}^{8}}+{}^{6}{{C}_{2}}{{x}^{7}}+\left( {}^{6}{{C}_{0}}-3{}^{6}{{C}_{6}} \right){{x}^{6}}-{}^{6}{{C}_{4}}12{{x}^{5}}-{}^{6}{{C}_{2}}{{x}^{4}}+3{}^{6}{{C}_{6}}{{x}^{3}}+{}^{6}{{C}_{4}}{{x}^{2}}-{}^{6}{{C}_{6}} \right)\] ---(4).
We know that the degree of the polynomial is defined as the highest power of independent variables present in that polynomial.
In equation (5), we can see that ‘x’ is the independent variable and the highest power of x is 9. This gives us the degree of the polynomial as 9.
∴ The degree of the polynomial ${{\left[ x+\sqrt{{{x}^{3}}-1} \right]}^{6}}+{{\left[ x-\sqrt{{{x}^{3}}-1} \right]}^{6}}$ is 9.
Note: We can also solve this problem by taking ${{x}^{\dfrac{3}{2}}}$ common from both the terms as shown below.
We have \[{{\left[ x+\sqrt{{{x}^{3}}-1} \right]}^{6}}+{{\left[ x-\sqrt{{{x}^{3}}-1} \right]}^{6}}={{\left[ x+\sqrt{{{x}^{3}}\left( 1-\dfrac{1}{{{x}^{3}}} \right)} \right]}^{6}}+{{\left[ x-\sqrt{{{x}^{3}}\left( 1-\dfrac{1}{{{x}^{3}}} \right)} \right]}^{6}}\].
We know that $\sqrt{{{a}^{m}}b}={{a}^{\dfrac{m}{2}}}\sqrt{b}$.
\[\Rightarrow {{\left[ x+\sqrt{{{x}^{3}}-1} \right]}^{6}}+{{\left[ x-\sqrt{{{x}^{3}}-1} \right]}^{6}}={{\left[ x+{{x}^{\dfrac{3}{2}}}\sqrt{\left( 1-\dfrac{1}{{{x}^{3}}} \right)} \right]}^{6}}+{{\left[ x-{{x}^{\dfrac{3}{2}}}\sqrt{\left( 1-\dfrac{1}{{{x}^{3}}} \right)} \right]}^{6}}\].
\[\Rightarrow {{\left[ x+\sqrt{{{x}^{3}}-1} \right]}^{6}}+{{\left[ x-\sqrt{{{x}^{3}}-1} \right]}^{6}}={{\left[ {{x}^{\dfrac{3}{2}}}\left( \dfrac{1}{{{x}^{\dfrac{1}{2}}}}+\sqrt{\left( 1-\dfrac{1}{{{x}^{3}}} \right)} \right) \right]}^{6}}+{{\left[ {{x}^{\dfrac{3}{2}}}\left( \dfrac{1}{{{x}^{\dfrac{1}{2}}}}-\sqrt{\left( 1-\dfrac{1}{{{x}^{3}}} \right)} \right) \right]}^{6}}\].
We know that ${{\left( ab \right)}^{m}}={{a}^{m}}{{b}^{m}}$.
\[\Rightarrow {{\left[ x+\sqrt{{{x}^{3}}-1} \right]}^{6}}+{{\left[ x-\sqrt{{{x}^{3}}-1} \right]}^{6}}={{\left( {{x}^{\dfrac{3}{2}}} \right)}^{6}}{{\left( \dfrac{1}{{{x}^{\dfrac{1}{2}}}}+\sqrt{\left( 1-\dfrac{1}{{{x}^{3}}} \right)} \right)}^{6}}+{{\left( {{x}^{\dfrac{3}{2}}} \right)}^{6}}{{\left( \dfrac{1}{{{x}^{\dfrac{1}{2}}}}-\sqrt{\left( 1-\dfrac{1}{{{x}^{3}}} \right)} \right)}^{6}}\].
\[\Rightarrow {{\left[ x+\sqrt{{{x}^{3}}-1} \right]}^{6}}+{{\left[ x-\sqrt{{{x}^{3}}-1} \right]}^{6}}=\left( {{x}^{9}} \right){{\left( \dfrac{1}{{{x}^{\dfrac{1}{2}}}}+\sqrt{\left( 1-\dfrac{1}{{{x}^{3}}} \right)} \right)}^{6}}+\left( {{x}^{9}} \right){{\left( \dfrac{1}{{{x}^{\dfrac{1}{2}}}}-\sqrt{\left( 1-\dfrac{1}{{{x}^{3}}} \right)} \right)}^{6}}\].
\[\Rightarrow {{\left[ x+\sqrt{{{x}^{3}}-1} \right]}^{6}}+{{\left[ x-\sqrt{{{x}^{3}}-1} \right]}^{6}}=\left( {{x}^{9}} \right)\left( {{\left( \dfrac{1}{{{x}^{\dfrac{1}{2}}}}+\sqrt{\left( 1-\dfrac{1}{{{x}^{3}}} \right)} \right)}^{6}}+{{\left( \dfrac{1}{{{x}^{\dfrac{1}{2}}}}-\sqrt{\left( 1-\dfrac{1}{{{x}^{3}}} \right)} \right)}^{6}} \right)\].
We can see that the power of x we get in the expansion will be less than or equal to 0, which makes 9 as the highest power of x as well as the degree of the polynomial.
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