Question

Expand the following expression in ascending powers of ‘x’ as far as ${{x}^{3}}$.
\[\dfrac{1}{1+ax-a{{x}^{2}}-{{x}^{3}}}\]

Answer 1

Hint: Suppose the expansion of the given term as \[\left( {{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+{{a}_{3}}{{x}^{3}}+..... \right)\]and compare the terms of variable ‘x’ in both sides of the equation to get number of equations. Solve them to get the values of\[{{a}_{0}},{{a}_{1}},{{a}_{2}},{{a}_{3}}\]. Don’t write the higher power terms as it is not required.

Complete step-by-step answer:

Given expression in the problem is
\[\dfrac{1}{1+ax-a{{x}^{2}}-{{x}^{3}}}........(i)\]
As, we need to expand the given expression in equation (i) to the powers of ‘x’. So, let us assume the expansion of the expression as
\[\dfrac{1}{1+ax-a{{x}^{2}}-{{x}^{3}}}={{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+{{a}_{3}}{{x}^{3}}+.....\]
On cross-multiplying the above equation, we get
\[1=\left( 1+ax-a{{x}^{2}}-{{x}^{3}} \right)\left( {{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+{{a}_{3}}{{x}^{3}}....... \right)\]
Now, let us multiply the brackets in Right-hand side, only upto the , as coefficient of higher powers is not asked, so we get
\[1=\left( {{a}_{0}}+{{a}_{0}}ax-{{a}_{0}}a{{x}^{2}}-{{a}_{0}}{{x}^{3}} \right)+\left( {{a}_{1}}x+a{{a}_{1}}{{x}^{2}}-a{{a}_{1}}{{x}^{3}} \right)+\left( {{a}_{2}}{{x}^{2}}+a{{a}_{2}}{{x}^{3}} \right)+\left( {{a}_{3}}{{x}^{3}} \right)+....\]
Now, taking the same powers of x in bracket, we get
\[\begin{align}
  & 1={{a}_{0}}+{{a}_{0}}ax+{{a}_{1}}x-{{a}_{0}}a{{x}^{2}}+a{{a}_{1}}{{x}^{2}}-{{a}_{2}}{{x}^{2}}-{{a}_{0}}{{x}^{3}}-a{{a}_{1}}{{x}^{3}}+a{{a}_{2}}{{x}^{3}}+{{a}_{3}}{{x}^{3}}+...... \\
 & \Rightarrow 1={{a}_{0}}+x\left( {{a}_{0}}a+{{a}_{1}} \right)+{{x}^{2}}\left( -{{a}_{0}}a+a{{a}_{1}}+{{a}_{2}} \right)+{{x}^{3}}\left( -{{a}_{0}}-a{{a}_{1}}+a{{a}_{2}}+{{a}_{3}} \right)+....... \\
\end{align}\]
Now, we can write the left hand side of the above equation a
\[1.{{x}^{0}}+0.x+0.{{x}^{2}}+0.{{x}^{3}}={{a}_{0}}+x\left( {{a}_{0}}a+{{a}_{1}} \right)+{{x}^{2}}\left( -{{a}_{0}}a+a{{a}_{1}}+{{a}_{2}} \right)+{{x}^{3}}\left( -{{a}_{0}}-a{{a}_{1}}+a{{a}_{2}}+{{a}_{3}} \right)...........\]
Now, compare the left hand side and right hand side of the above equation and get equations as
\[\begin{align}
  & {{a}_{0}}=1.......(ii) \\
 & {{a}_{0}}a+{{a}_{1}}=0......(iii) \\
 & -{{a}_{0}}a+a{{a}_{1}}+{{a}_{2}}=0......(iv) \\
 & -{{a}_{0}}-a{{a}_{1}}+a{{a}_{2}}+{{a}_{3}}=0.......(v) \\
\end{align}\]
So, we get value of as
\[{{a}_{0}}=1\]
Put value of \[{{a}_{0}}\]from equation (ii) to equation (iii) as
\[\begin{align}
  & 1(a)+{{a}_{1}}=0 \\
 & {{a}_{1}}=-a.......(vi) \\
\end{align}\]
Now, put value of\[{{a}_{0}}{{a}_{1}}\]to the equation (iv) to get value of \[{{a}_{2}}\]as
\[\begin{align}
  & -1(a)+a(-a)+{{a}_{2}}=0 \\
 & -a-{{a}^{_{2}}}+{{a}_{2}}=0 \\
 & {{a}_{2}}={{a}^{_{2}}}+a.......(vii) \\
\end{align}\]
Now, we can put value of \[{{a}_{0}},{{a}_{1}},{{a}_{2}}\]to equation (v) as
\[\begin{align}
  & -1-a(-a)+a({{a}^{2}}+a)+{{a}_{3}}=0 \\
 & -1+{{a}^{2}}+{{a}^{3}}+{{a}^{2}}+{{a}_{3}}=0 \\
 & {{a}_{3}}=1-2{{a}^{2}}-{{a}^{3}}........(viii) \\
\end{align}\]
Hence, we can write the expansion of the given expression as
\[\dfrac{1}{1+a{{x}^{2}}-a{{x}^{2}}-{{x}^{3}}}=1-ax+({{a}^{2}}+a){{x}^{2}}+(1-2{{a}^{2}}-{{a}^{3}}){{x}^{3}}+..........\]

Note: One may use the binomial expansion of any index given as
\[{{(1+x)}^{n}}=1+nx+\dfrac{n(n-1)}{2!}{{x}^{2}}+\dfrac{n(n-1)(n-2)}{3!}{{x}^{3}}+.........\]
It can be applied as
\[{{\left( 1+ax-a{{x}^{2}}-{{x}^{3}} \right)}^{-1}}=1+\left( -1 \right)\left( ax-a{{x}^{2}}-{{x}^{3}} \right)+\dfrac{\left( -1 \right)\left( -1-1 \right)}{2!}{{\left( ax-a{{x}^{2}}-{{x}^{3}} \right)}^{2}}+\dfrac{\left( -1 \right)\left( -1-1 \right)\left( -1-2 \right)}{3!}{{\left( ax-a{{x}^{2}}-{{x}^{3}} \right)}^{3}}+......\]
Now, use the above approach as well to expand the given series.
Calculation is the key point of the question. We don’t need to take the terms with higher powers than\[{{x}^{3}}\].