Question

If \[a = \sum\limits_{n = 0}^\infty {\dfrac{{{x^{3n}}}}{{3n!}}} ,b = \sum\limits_{n = 1}^\infty {\dfrac{{{x^{3n - 2}}}}{{\left( {3n - 2} \right)!}},c = \sum\limits_{n = 1}^\infty {\dfrac{{{x^{3n - 1}}}}{{\left( {3n - 1} \right)!}}} } \] Then the value of \[{a^3} + {b^3} + {c^3} - 3abc\]is
(A) \[1\]
(B) \[0\]
(C) \[ - 1\]
(D) \[2\]

Answer 1

Hint: In this question, we have to fins the value of \[{a^3} + {b^3} + {c^3} - 3abc\].Firstly, we will expand the summation \[a\] to \[n = 1\] and add the given summations and then expand them by substituting the values. Further, we will find \[a + b\omega + c{\omega ^2}\] and \[a + b{\omega ^2} + c\omega \]. Then we will multiply all these equations to find the value of \[{a^3} + {b^3} + {c^3} - 3abc\]

Formula used:
The formulas here we used to solve this question is,
1. \[{e^x} = 1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + ....\]
2. \[{a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left( {a + b\omega + c{\omega ^2}} \right)\left( {a + b{\omega ^2} + c\omega } \right)\] where \[\omega \] is complex cube root of unity.

Complete step-by-step solution:
The given terms are \[a = \sum\limits_{n = 0}^\infty {\dfrac{{{x^{3n}}}}{{3n!}}} ,b = \sum\limits_{n = 1}^\infty {\dfrac{{{x^{3n - 2}}}}{{\left( {3n - 2} \right)!}},c = \sum\limits_{n = 1}^\infty {\dfrac{{{x^{3n - 1}}}}{{\left( {3n - 1} \right)!}}} } \]
Firstly, we will expand the summation \[a\] to \[n = 1\] by substituting \[n = 0\], we get
\[\begin{array}{l}a = \dfrac{{{x^{3 \times 0}}}}{{3 \times 0}} + \sum\limits_{n = 1}^\infty {\dfrac{{{x^{3n}}}}{{3n!}}} \\a = 0 + \sum\limits_{n = 1}^\infty {\dfrac{{{x^{3n}}}}{{3n!}}} \\a = \sum\limits_{n = 1}^\infty {\dfrac{{{x^{3n - 3}}}}{{\left( {3n - 3} \right)!}}} \end{array}\]
Now, we will add all the terms as \[a + b + c\], we get
\[a + b + c = \sum\limits_{n = 1}^\infty {\dfrac{{{x^{3n - 3}}}}{{\left( {3n - 3} \right)!}}} + \sum\limits_{n = 1}^\infty {\dfrac{{{x^{3n - 2}}}}{{\left( {3n - 2} \right)!}} + \sum\limits_{n = 1}^\infty {\dfrac{{{x^{3n - 1}}}}{{\left( {3n - 1} \right)!}}} } \]
Further, we will expand all the terms by substituting \[n = 1\], we get
\[a + b + c = 1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + ....\] (1)
As we know that \[{e^x} = 1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + ....\].
So, substitute this in equation (1), we get
\[a + b + c = {e^x}\] ……(2)
Further, we will find \[a + b\omega + c{\omega ^2}\] by comparing \[x\] with \[\omega x\] in equation (1), we get
\[a + b\omega + c{\omega ^2} = 1 + \dfrac{{\omega x}}{{1!}} + \dfrac{{{\omega ^2}{x^2}}}{{2!}} + \dfrac{{{\omega ^3}{x^3}}}{{3!}} + ....\]
Now, we will again use this formula \[{e^x} = 1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + ....\], we get
\[a + b\omega + c{\omega ^2} = {e^{\omega x}}\] (3)
Further, we will find \[a + b{\omega ^2} + c\omega \] by comparing \[x\] with \[{\omega ^2}x\] in equation (1), we get
\[a + b{\omega ^2} + c\omega = 1 + \dfrac{{{\omega ^2}x}}{{1!}} + \dfrac{{{\omega ^3}{x^2}}}{{2!}} + \dfrac{{{\omega ^4}{x^3}}}{{3!}} + ....\]
Furthermore, we will again use this formula \[{e^x} = 1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + ....\], we get
\[a + b{\omega ^2} + c\omega = {e^{{\omega ^2}x}}\] (4)
Now, we will find the value of \[{a^3} + {b^3} + {c^3} - 3abc\] by applying the formula \[{a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left( {a + b\omega + c{\omega ^2}} \right)\left( {a + b{\omega ^2} + c\omega } \right)\] and substitute the value from equation (1), (2) and (3), we get
\[{a^3} + {b^3} + {c^3} - 3abc = {e^x} \times {e^{\omega x}} \times {e^{{\omega ^2}x}}\]
Further, we will taking out \[{e^x}\] as common, we get
\[{a^3} + {b^3} + {c^3} - 3abc = {e^{x\left( {1 + \omega + {\omega ^2}} \right)}}\]
As we know that \[\left( {1 + \omega + {\omega ^2}} \right) = 0\].
So, substitute this in above, we get
\[\begin{array}{l}{a^3} + {b^3} + {c^3} - 3abc = {e^{x\left( 0 \right)}}\\{a^3} + {b^3} + {c^3} - 3abc = {e^0}\\{a^3} + {b^3} + {c^3} - 3abc = 1\end{array}\]

Hence Option A) is the correct answer.

Note : In this type of questions, we should now how to solve summation by substituting terms and also know the expansion of terms and algebraic identities. We should also remember the equations and values of complex cube root of unity.