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Hint: In this question, we first need to write the sum, sum of product of two roots at a time, product of the roots in terms of the coefficients of the given cubic polynomial. Then solve the equations formed in terms of a and b to get their respective values.

Complete step by step answer:

CUBIC POLYNOMIAL: A cubic polynomial is a polynomial having degree three.

The general form of a cubic polynomial is given by \[{{a}_{0}}{{x}^{3}}+{{a}_{1}}{{x}^{2}}+{{a}_{2}}x+{{a}_{3}}=0\]

ZERO OF A POLYNOMIAL: A real number \[\alpha \] is a zero of the polynomial p(x), if and only if \[p\left( \alpha \right)=0\]

Now, let us assume the zeroes of the cubic polynomial as

\[\alpha ,\beta ,\gamma \]

Now, from the cubic polynomial \[{{a}_{0}}{{x}^{3}}+{{a}_{1}}{{x}^{2}}+{{a}_{2}}x+{{a}_{3}}=0\] we have,

Sum of the roots of the cubic polynomial is given by

\[\alpha +\beta +\gamma =\dfrac{-{{a}_{1}}}{{{a}_{0}}}\]

Now, sum of the product of the two roots at a time is given by

\[\alpha \beta +\beta \gamma +\gamma \alpha =\dfrac{{{a}_{2}}}{{{a}_{0}}}\]

Now, product of all the roots is given by

\[\alpha \beta \gamma =\dfrac{-{{a}_{3}}}{{{a}_{0}}}\]

Now, on comparing the zeroes given in the question to the formulae above we get,

\[\alpha =a-b,\beta =a,\gamma =a+b\]

Now, again on comparing the given cubic polynomial with the general form we get,

\[\Rightarrow {{x}^{3}}-3{{x}^{2}}+x+1=0\]

\[{{a}_{0}}=1,{{a}_{1}}=-3,{{a}_{2}}=1,{{a}_{3}}=1\]

Now, from the sum of the roots formula on substituting the respective values we get,

\[\Rightarrow a-b+a+a+b=\dfrac{-\left( -3 \right)}{1}\]

Now, this can be further written as

\[\Rightarrow a+a+a=3\]

Now, on further simplification we get,

\[\Rightarrow 3a=3\]

Now, on dividing with 3 on both sides we get,

\[\therefore a=1\]

Let us now consider the product of the roots formula and substitute the respective values.

\[\Rightarrow \left( a-b \right)\times a\times \left( a+b \right)=\dfrac{-1}{1}\]

Now, on substituting the value of a this can be further written as

\[\Rightarrow \left( 1-b \right)\times 1\times \left( 1+b \right)=-1\]

Now, on further simplification we get,

\[\Rightarrow 1-{{b}^{2}}=-1\]

Now, on rearranging the terms we get,

\[\Rightarrow {{b}^{2}}=2\]

\[\therefore b=\sqrt{2}\]

Note:

Instead of using the formulae of sum of the roots and product of the roots we can also use the sum of the product of two roots taken at a time and sum of the roots. Both the methods give the same result.

Here, we considered the value of b only positive because as already given that the roots are \[a-b,a,a+b\] we get the same values because the value of b is irrational. That means the roots will be conjugate so either we consider the negative value or positive value the roots will be the same for the given cubic polynomial.

Complete step by step answer:

CUBIC POLYNOMIAL: A cubic polynomial is a polynomial having degree three.

The general form of a cubic polynomial is given by \[{{a}_{0}}{{x}^{3}}+{{a}_{1}}{{x}^{2}}+{{a}_{2}}x+{{a}_{3}}=0\]

ZERO OF A POLYNOMIAL: A real number \[\alpha \] is a zero of the polynomial p(x), if and only if \[p\left( \alpha \right)=0\]

Now, let us assume the zeroes of the cubic polynomial as

\[\alpha ,\beta ,\gamma \]

Now, from the cubic polynomial \[{{a}_{0}}{{x}^{3}}+{{a}_{1}}{{x}^{2}}+{{a}_{2}}x+{{a}_{3}}=0\] we have,

Sum of the roots of the cubic polynomial is given by

\[\alpha +\beta +\gamma =\dfrac{-{{a}_{1}}}{{{a}_{0}}}\]

Now, sum of the product of the two roots at a time is given by

\[\alpha \beta +\beta \gamma +\gamma \alpha =\dfrac{{{a}_{2}}}{{{a}_{0}}}\]

Now, product of all the roots is given by

\[\alpha \beta \gamma =\dfrac{-{{a}_{3}}}{{{a}_{0}}}\]

Now, on comparing the zeroes given in the question to the formulae above we get,

\[\alpha =a-b,\beta =a,\gamma =a+b\]

Now, again on comparing the given cubic polynomial with the general form we get,

\[\Rightarrow {{x}^{3}}-3{{x}^{2}}+x+1=0\]

\[{{a}_{0}}=1,{{a}_{1}}=-3,{{a}_{2}}=1,{{a}_{3}}=1\]

Now, from the sum of the roots formula on substituting the respective values we get,

\[\Rightarrow a-b+a+a+b=\dfrac{-\left( -3 \right)}{1}\]

Now, this can be further written as

\[\Rightarrow a+a+a=3\]

Now, on further simplification we get,

\[\Rightarrow 3a=3\]

Now, on dividing with 3 on both sides we get,

\[\therefore a=1\]

Let us now consider the product of the roots formula and substitute the respective values.

\[\Rightarrow \left( a-b \right)\times a\times \left( a+b \right)=\dfrac{-1}{1}\]

Now, on substituting the value of a this can be further written as

\[\Rightarrow \left( 1-b \right)\times 1\times \left( 1+b \right)=-1\]

Now, on further simplification we get,

\[\Rightarrow 1-{{b}^{2}}=-1\]

Now, on rearranging the terms we get,

\[\Rightarrow {{b}^{2}}=2\]

\[\therefore b=\sqrt{2}\]

Note:

Instead of using the formulae of sum of the roots and product of the roots we can also use the sum of the product of two roots taken at a time and sum of the roots. Both the methods give the same result.

Here, we considered the value of b only positive because as already given that the roots are \[a-b,a,a+b\] we get the same values because the value of b is irrational. That means the roots will be conjugate so either we consider the negative value or positive value the roots will be the same for the given cubic polynomial.

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