Miscellaneous Exercise Binomial Theorme Solution - NCERT Class 11


Chapter 8 Ex.8.ME Question 1

Find a, b and n in the expansion of \({\left( {a + b} \right)^n}\) if the first three terms of the expansion are 729, 7290 and 30375 respectively.

 

Solution

Video Solution

 

It is known that \({\left( {r + 1} \right)^{th}}\) term, \(\left( {{T_{r + 1}}} \right)\)the binomial expression of \({\left( {a + b} \right)^n}\)is given by

\({T_{r + 1}} = {}^n{C_r}{a^{n - r}}{b^r}\)

The first three terms of expansion are 729, 7290 and 30375 respectively.

\[\begin{align}{T_1} &= {}^n{C_0}{a^{n - 0}}{b^0}\\&= {a^n}\\&= 729 \qquad \quad \ldots \left( 1 \right)\\\\{T_2} &= {}^n{C_1}{a^{n - 1}}{b^1}\\&= n{a^{n - 1}}b\\&= 7290\qquad \quad \ldots \left( 2 \right)\\\\{T_3} &= {}^n{C_1}{a^{n - 2}}{b^2}\\&= \frac{{n\left( {n - 1} \right)}}{2}{a^{n - 2}}{b^2}\\&= 30375\qquad \quad \ldots \left( 3 \right)\end{align}\]

Dividing (2) by (1), we obtain

\[\begin{align}\frac{{n{a^{n - 1}}b}}{{{a^n}}} &= \frac{{7290}}{{729}}\\\frac{{nb}}{a} &= 10 \qquad \quad \ldots (4)\end{align}\]

Dividing (3) by (2), we obtain

\[\begin{align}\frac{{n\left( {n - 1} \right){a^{n - 2}}{b^2}}}{{2n{a^{n - 1}}b}} &= \frac{{30375}}{{7290}}\\\frac{{\left( {n - 1} \right)b}}{{2a}} &= \frac{{30375}}{{7290}}\\\frac{{\left( {n - 1} \right)b}}{a} &= \frac{{30375 \times 2}}{{7290}}\\\frac{{nb}}{a} - \frac{b}{a} &= \frac{{25}}{3}\end{align}\]

\[\begin{align}10 - \frac{b}{a} &= \frac{{25}}{3} \qquad \quad \ldots \left( {{\rm{from}}\;{\rm{4}}} \right)\\\frac{b}{a} &= 10 - \frac{{25}}{3}\\\frac{b}{a} &= \frac{5}{3} \qquad \quad \ldots \left( 5 \right)\end{align}\]

Putting \(\frac{b}{a} = \frac{5}{3}\)in (4), we obtain

\[\begin{align}n \times \frac{5}{3} &= 10\\n &= 6\end{align}\]

Substituting \(n = 6\) in (1), we obtain

\[\begin{align}{a^6} &= 729\\a &= \sqrt[6]{{729}}\\&= 3\end{align}\]

From (5), we obtain

\[\begin{align}\frac{b}{3} &= \frac{5}{3}\\b &= 5\end{align}\]

Therefore, \(a = 3,b = 5,n = 6\)

Chapter 8 Ex.8.ME Question 2

Find \(a\) if the coefficients of \({x^2}\)and \({x^3}\) in the expansion of \({\left( {3 + ax} \right)^9}\) are equal.

 

Solution

Video Solution

 

It is known that \({\left( {r + 1} \right)^{th}}\) term, \(\left( {{T_{r + 1}}} \right)\)the binomial expression of \({\left( {a + b} \right)^n}\)is given by

\({T_{r + 1}} = {}^n{C_r}{a^{n - r}}{b^r}\)

Assuming that \({x^2}\) occurs in the \({\left( {r + 1} \right)^{th}}\) term of the expansion \({\left( {3 + ax} \right)^9}\) , we obtain

\[\begin{align}{T_{r + 1}} &= {}^9{C_r}{\left( 3 \right)^{9 - r}}{\left( {ax} \right)^r}\\&= {}^9{C_r}{\left( 3 \right)^{9 - r}}{a^r}{x^r}\end{align}\]

Comparing the indices of \(x\) in \({x^2}\)and in \(\left( {{T_{r + 1}}} \right)\), we obtain \(r = 2\)

Thus, the coefficient of \({{x}^{2}}\)is

\[\begin{align}{}^9{C_2}{\left( 3 \right)^{9 - 2}}{a^2} = \frac{{9!}}{{2!7!}}{\left( 3 \right)^7}{a^2}\\ = 36{\left( 3 \right)^7}{a^2}\end{align}\]

Assuming that \({x^3}\) occurs in the \({\left( {k + 1} \right)^{th}}\) term of the expansion \({\left( {3 + ax} \right)^9}\) , we obtain

\[\begin{align}{T_{k + 1}} &= {}^9{C_k}{\left( 3 \right)^{9 - k}}{\left( {ax} \right)^k}\\&= {}^9{C_k}{\left( 3 \right)^{9 - k}}{a^k}{x^k}\end{align}\]

Comparing the indices in \({x^2}\) and \({{x}^{3}}\) in \(\left( {{T}_{r+1}} \right)\), we obtain \(k=3\)

Thus, the coefficient of \({{x}^{3}}\)is

\[\begin{align}{}^9{C_3}{\left( 3 \right)^{9 - 3}}{a^3} &= \frac{{9!}}{{3!6!}}{\left( 3 \right)^6}{a^3}\\&= 84{\left( 3 \right)^6}{a^3}\end{align}\]

It is given that the coefficient of \({x^2}\) and \({x^3}\) are equal.

\[\begin{align}84{\left( 3 \right)^6}{a^3} &= 36{\left( 3 \right)^7}{a^2}\\84a &= 36 \times 3\\a &= \frac{{36 \times 3}}{{84}}\\&= \frac{9}{7}\end{align}\]

Hence, the required value of \(a = \frac{9}{7}\)

Chapter 8 Ex.8.ME Question 3

Find the coefficient of \({x^5}\)in the product \({\left( {1 + 2x} \right)^6}{\left( {1 - x} \right)^7}\)using binomial theorem.

 

Solution

Video Solution

 

Using binomial theorem, \({\left( {1 + 2x} \right)^6}\) and \({\left( {1 - x} \right)^7}\)can be expanded as

\[\begin{align}{\left( {1 + 2x} \right)^6} &= {}^6{C_0} + {}^6{C_1}\left( {2x} \right) + {}^6{C_2}{\left( {2x} \right)^2} + {}^6{C_3}{\left( {2x} \right)^3} + {}^6{C_4}{\left( {2x} \right)^4} + {}^6{C_5}{\left( {2x} \right)^5} + {}^6{C_6}{\left( {2x} \right)^6}\\&= 1 + 6\left( {2x} \right) + 15{\left( {2x} \right)^2} + 20{\left( {2x} \right)^3} + 15{\left( {2x} \right)^4} + 6{\left( {2x} \right)^5} + {\left( {2x} \right)^6}\\&= 1 + 12x + 60{x^2} + 160{x^3} + 240{x^4} + 192{x^5} + 64{x^6}\\\\{\left( {1 - x} \right)^7} &= {}^7{C_0} - {}^7{C_1}\left( x \right) + {}^7{C_2}{\left( x \right)^2} - {}^7{C_3}{\left( x \right)^3} + {}^7{C_4}{\left( x \right)^4} - {}^7{C_5}{\left( x \right)^5} + {}^7{C_6}{\left( x \right)^6} - {}^7{C_7}{\left( x \right)^7}\\&= 1 - 7x + 21{x^2} - 35{x^3} + 35{x^4} - 21{x^5} + 7{x^6} - {x^7}\end{align}\]

Therefore,

\[\begin{align}{\left( {1 + 2x} \right)^6}{\left( {1 - x} \right)^7} &= \left( {1 + 12x + 60{x^2} + 160{x^3} + 240{x^4} + 192{x^5} + 64{x^6}} \right)\left( {1 - 7x + 21{x^2} - 35{x^3} + 35{x^4} - 21{x^5} + 7{x^6} - {x^7}} \right)\\&= 1\left( { - 21{x^5}} \right) + \left( {12x} \right)\left( {35{x^4}} \right) + \left( {60{x^2}} \right)\left( { - 35{x^3}} \right) + \left( {160{x^3}} \right)\left( {21{x^2}} \right) + \left( {240{x^4}} \right)\left( { - 7x} \right) + \left( {192{x^5}} \right)\left( 1 \right)\\&= 171{x^5}\end{align}\]

Thus, the coefficient of \({x^5}\)in the product \({\left( {1 + 2x} \right)^6}{\left( {1 - x} \right)^7}\) is 171.

Chapter 8 Ex.8.ME Question 4

If a and b are distinct integers, prove that \(a - b\) is a factor of \({a^n} - {b^n}\) whenever \(n\) is a positive integer.

[Hint: write \({a^n} = {\left( {a-b + b} \right)^n}\) and expand]

 

Solution

Video Solution

 

In order to prove that \(a - b\) is a factor of \({a^n} - {b^n}\), we have to prove that \({a^n} - {b^n} = k\left( {a - b} \right)\) where k is some natural number.

We can write \(a\) as

\[\begin{align}a &= a - b + b\\{a^n} &= {\left( {a - b + b} \right)^n}\\{a^n} &= {\left[ {\left( {a - b} \right) + b} \right]^n}\\{a^n} &= {}^n{C_0}{\left( {a - b} \right)^n} + {}^n{C_1}{\left( {a - b} \right)^{n - 1}}b + \ldots + {}^n{C_{n - 1}}\left( {a - b} \right){b^{n - 1}} + {}^n{C_n}{b^n}\\{a^n} &= {\left( {a - b} \right)^n} + {}^n{C_1}{\left( {a - b} \right)^{n - 1}}b + \ldots + {}^n{C_{n - 1}}\left( {a - b} \right){b^{n - 1}} + {b^n}\\{a^n} - {b^n} &= \left( {a - b} \right)\left[ {{{\left( {a - b} \right)}^{n - 1}} + {}^n{C_1}{{\left( {a - b} \right)}^{n - 2}}b + \ldots + {}^n{C_{n - 1}}{b^{n - 1}}} \right]\\{a^n} - {b^n} &= k\left( {a - b} \right)\end{align}\]

where \(k = \left[ {{{\left( {a - b} \right)}^{n - 1}} + {}^n{C_1}{{\left( {a - b} \right)}^{n - 2}}b + \ldots + {}^n{C_{n - 1}}{b^{n - 1}}} \right]\)

This shows that \(a - b\) is a factor of \({a^n} - {b^n}\) whenever \(n\) is a positive integer.

Hence proved.

Chapter 8 Ex.8.ME Question 5

Question 5:

Evaluate \({\left( {\sqrt 3 + \sqrt 2 } \right)^6} - {\left( {\sqrt 3 - \sqrt 2 } \right)^6}\)

 

Solution

Video Solution

 

Using binomial theorem,

\[\begin{align}{\left( {a + b} \right)^6} &= {}^6{C_0}{a^6} + {}^6{C_1}{a^5}b + {}^6{C_2}{a^4}{b^2} + {}^6{C_3}{a^3}{b^3} + {}^6{C_4}{a^2}{b^4} + {}^6{C_5}a{b^5} + {}^6{C_6}{b^6}\\&= {a^6} + 6{a^5}b + 15{a^4}{b^2} + 20{a^3}{b^3} + 15{a^2}{b^4} + 6a{b^5} + {b^6}\\\\{\left( {a - b} \right)^6} &= {}^6{C_0}{a^6} - {}^6{C_1}{a^5}b + {}^6{C_2}{a^4}{b^2} - {}^6{C_3}{a^3}{b^3} + {}^6{C_4}{a^2}{b^4} - {}^6{C_5}a{b^5} + {}^6{C_6}{b^6}\\&= {a^6} - 6{a^5}b + 15{a^4}{b^2} - 20{a^3}{b^3} + 15{a^2}{b^4} - 6a{b^5} + {b^6}\end{align}\]

Therefore,

\({\left( {a + b} \right)^6} - {\left( {a - b} \right)^6} = 2\left[ {6{a^5}b + 20{a^3}{b^3} + 6a{b^5}} \right]\)

Putting \(a = \sqrt 3 \) and \(b = \sqrt 2 \), we obtain,

\[\begin{align}{\left( {\sqrt 3 + \sqrt 2 } \right)^6} - {\left( {\sqrt 3 - \sqrt 2 } \right)^6} &= 2\left[ {6{{\left( {\sqrt 3 } \right)}^5}\left( {\sqrt 2 } \right) + 20{{\left( {\sqrt 3 } \right)}^3}{{\left( {\sqrt 2 } \right)}^3} + 6\left( {\sqrt 3 } \right){{\left( {\sqrt 2 } \right)}^5}} \right]\\&= 2\left[ {54\sqrt 6 + 120\sqrt 6 + 24\sqrt 6 } \right]\\&= 2 \times 198\sqrt 6 \\&= 396\sqrt 6 \end{align}\]

Chapter 8 Ex.8.ME Question 6

Find the value of \({\left( {{a^2} + \sqrt {{a^2} - 1} } \right)^4} + {\left( {{a^2} - \sqrt {{a^2} - 1} } \right)^4}\)

 

Solution

Video Solution

 

Using binomial theorem,

\[\begin{align}{\left( {x + y} \right)^4} &= {}^4{C_0}{x^4} + {}^4{C_1}{x^3}y + {}^4{C_2}{x^2}{y^2} + {}^4{C_3}x{y^3} + {}^4{C_4}{y^4}\\&= {x^4} + 4{x^3}y + 6{x^2}{y^2} + 4x{y^3} + {y^4}\\\\{\left( {x - y} \right)^4} &= {}^4{C_0}{x^4} - {}^4{C_1}{x^3}y + {}^4{C_2}{x^2}{y^2} - {}^4{C_3}x{y^3} + {}^4{C_4}{y^4}\\&= {x^4} - 4{x^3}y + 6{x^2}{y^2} - 4x{y^3} + {y^4}\end{align}\]

Therefore,

\({\left( {x + y} \right)^4} + {\left( {x - y} \right)^4} = 2\left( {{x^4} + 6{x^2}{y^2} + {y^4}} \right)\)

Putting \(x = {a^2}\) and \(y = \sqrt {{a^2} - 1} \), we obtain

\[\begin{align}{\left( {{a^2} + \sqrt {{a^2} - 1} } \right)^4} + {\left( {{a^2} - \sqrt {{a^2} - 1} } \right)^4} &= 2\left[ {{{\left( {{a^2}} \right)}^4} + 6{{\left( {{a^2}} \right)}^2}{{\left( {\sqrt {{a^2} - 1} } \right)}^2} + {{\left( {\sqrt {{a^2} - 1} } \right)}^4}} \right]\\&= 2\left[ {{a^8} + 6{a^4}\left( {{a^2} - 1} \right) + {{\left( {{a^2} - 1} \right)}^2}} \right]\\&= 2\left[ {{a^8} + 6{a^6} - 6{a^4} + {a^4} - 2{a^2} + 1} \right]\\&= 2\left[ {{a^8} + 6{a^6} - 5{a^4} - 2{a^2} + 1} \right]\\&= 2{a^8} + 12{a^6} - 10{a^4} - 4{a^2} + 2\end{align}\]

Chapter 8 Ex.8.ME Question 7

Find an approximation of \({\left( {0.99} \right)^5}\)using the first three terms of its expansion.

 

Solution

Video Solution

 

We can express \(0.99\) as the difference of two numbers whose powers are easier to calculate.

Hence, \(0.99 = 1 - 0.01\)

Therefore,

\[\begin{align}{\left( {0.99} \right)^5} &= {\left( {1 - 0.01} \right)^5}\\&= {}^5{C_0}{\left( 1 \right)^5} - {}^5{C_1}{\left( 1 \right)^4}\left( {0.01} \right) + {}^5{C_2}{\left( 1 \right)^3}{\left( {0.01} \right)^2}\\&= 1 - 5\left( {0.01} \right) + 10{\left( {0.01} \right)^2}\\&= 1 - 0.05 + 0.001\\&= 1.001 - 0.05\\&= 0.951\end{align}\]

Hence, the approximation of \({\left( {0.99} \right)^5}\) is \(0.951\)

Chapter 8 Ex.8.ME Question 8

Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of \({\left( {\sqrt[4]{2} + \frac{1}{{\sqrt[4]{3}}}} \right)^n}\) is \(\sqrt 6 :1\)

 

Solution

Video Solution

 

In the expansion,

\({\left( {a + b} \right)^n} = {}^n{C_0}{a^n} + {}^n{C_1}{a^{n - 1}}b + {}^n{C_2}{a^{n - 2}}{b^2} + ................ + {}^n{C_{n - 1}}a{b^n}^{ - 1} + {}^n{C_n}{b^n}\)

Fifth term from the beginning is \({}^n{C_4}{a^{n - 4}}{b^4}\)

Fifth term from the end is \({}^n{C_4}{a^4}{b^{n - 4}}\)

Therefore, in the expansion of \({\left( {\sqrt[4]{2} + \frac{1}{{\sqrt[4]{3}}}} \right)^n}\);

fifth term from the beginning is \({}^n{C_4}{\left( {\sqrt[4]{2}} \right)^{n - 4}}{\left( {\frac{1}{{\sqrt[4]{3}}}} \right)^4}\)and

fifth term from the end is \({}^n{C_{n - 4}}{\left( {\sqrt[4]{2}} \right)^4}{\left( {\frac{1}{{\sqrt[4]{3}}}} \right)^{n - 4}}\)

\[\begin{align}{}^n{C_4}{\left( {\sqrt[4]{2}} \right)^{n - 4}}{\left( {\frac{1}{{\sqrt[4]{3}}}} \right)^4} &= {}^n{C_4}\frac{{{{\left( {\sqrt[4]{2}} \right)}^n}}}{{{{\left( {\sqrt[4]{2}} \right)}^4}}} \cdot \frac{1}{3}\\&= \frac{{n!}}{{6 \cdot 4!\left( {n - 4} \right)!}}{\left( {\sqrt[4]{2}} \right)^n} \qquad \quad \ldots \left( 1 \right)\\\\{}^n{C_{n - 4}}{\left( {\sqrt[4]{2}} \right)^4}{\left( {\frac{1}{{\sqrt[4]{3}}}} \right)^{n - 4}} = {}^n{C_{n - 4}}\frac{{{{\left( {\sqrt[4]{3}} \right)}^4}}}{{{{\left( {\sqrt[4]{3}} \right)}^n}}}\\&= {}^n{C_{n - 4}} \cdot 2 \cdot \frac{3}{{{{\left( {\sqrt[4]{3}} \right)}^n}}}\\&= \frac{{6n!}}{{\left( {n - 4} \right)!4!}} \cdot \frac{1}{{{{\left( {\sqrt[4]{3}} \right)}^n}}}\qquad \quad \ldots \left( 2 \right)\end{align}\]

It is given that the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of \({\left( {\sqrt[4]{2} + \frac{1}{{\sqrt[4]{3}}}} \right)^n}\)is \(\sqrt 6 :1\)

Therefore,

\[\begin{align}\left[ {\frac{{n!}}{{6 \cdot 4!\left( {n - 4} \right)!}}{{\left( {\sqrt[4]{2}} \right)}^n}} \right]:\left[ {\frac{{6n!}}{{\left( {n - 4} \right)!4!}} \cdot \frac{1}{{{{\left( {\sqrt[4]{3}} \right)}^n}}}} \right] &= \sqrt 6 :1\\\frac{{{{\left( {\sqrt[4]{2}} \right)}^n}}}{6}:\frac{6}{{{{\left( {\sqrt[4]{3}} \right)}^n}}} &= \sqrt 6 :1\\\frac{{{{\left( {\sqrt[4]{2}} \right)}^n}}}{6} \times \frac{{{{\left( {\sqrt[4]{3}} \right)}^n}}}{6} &= \sqrt 6 \\{\left( {\sqrt[4]{6}} \right)^n} = 36\sqrt 6 \\{\left( 6 \right)^{\frac{n}{4}}} &= {\left( 6 \right)^{\frac{5}{2}}}\\\frac{n}{4} &= \frac{5}{2}\\n &= 4 \times \frac{5}{2}\\&= 10\end{align}\]

Thus the value of \(n = 10\).

Chapter 8 Ex.8.ME Question 9

Expand using binomial theorem \({\left( {1 + \frac{x}{2} - \frac{2}{x}} \right)^4},x \ne 0\)

 

Solution

Video Solution

 

\[\begin{align}{\left( {1 + \frac{x}{2} - \frac{2}{x}} \right)^4} &= {}^n{C_0}{\left( {1 + \frac{x}{2}} \right)^4} - {}^n{C_1}{\left( {1 + \frac{x}{2}} \right)^3}\left( {\frac{2}{x}} \right) + {}^n{C_2}{\left( {1 + \frac{x}{2}} \right)^2}{\left( {\frac{2}{x}} \right)^2} - {}^n{C_3}\left( {1 + \frac{x}{2}} \right){\left( {\frac{2}{x}} \right)^3} + {}^n{C_4}{\left( {\frac{2}{x}} \right)^4}\\&= {\left( {1 + \frac{x}{2}} \right)^4} - 4{\left( {1 + \frac{x}{2}} \right)^3}\left( {\frac{2}{x}} \right) + 6\left( {1 + x + \frac{{{x^2}}}{4}} \right)\left( {\frac{4}{{{x^2}}}} \right) - 4\left( {1 + \frac{x}{2}} \right)\left( {\frac{8}{{{x^3}}}} \right) + \frac{{16}}{{{x^4}}}\\&= {\left( {1 + \frac{x}{2}} \right)^4} - \frac{8}{x}{\left( {1 + \frac{x}{2}} \right)^3} + \frac{{24}}{{{x^2}}} + \frac{{24}}{x} + 6 - \frac{{32}}{{{x^3}}} - \frac{{16}}{{{x^2}}} + \frac{{16}}{{{x^4}}}\\&= {\left( {1 + \frac{x}{2}} \right)^4} - \frac{8}{x}{\left( {1 + \frac{x}{2}} \right)^3} + \frac{8}{{{x^2}}} + \frac{{24}}{x} + 6 - \frac{{32}}{{{x^3}}} + \frac{{16}}{{{x^4}}} \qquad  \ldots \left( 1 \right)\end{align}\]

Again, by using binomial theorem, we obtain

\[\begin{align}{\left( {1 + \frac{x}{2}} \right)^4} &= {}^4{C_0}{\left( 1 \right)^4} + {}^4{C_1}{\left( 1 \right)^3}\left( {\frac{x}{2}} \right) + {}^4{C_2}{\left( 1 \right)^2}{\left( {\frac{x}{2}} \right)^2} + {}^4{C_3}\left( 1 \right){\left( {\frac{x}{2}} \right)^3} + {}^4{C_4}{\left( {\frac{x}{2}} \right)^4}\\ &= 1 + 4 \times \frac{x}{2} + 6 \times \frac{{{x^4}}}{4} + 4 \times \frac{{{x^3}}}{8} + \frac{{{x^4}}}{{16}}\\ &= 1 + 2x + \frac{{3{x^2}}}{2} + \frac{{{x^3}}}{2} + \frac{{{x^4}}}{{16}} \qquad \ldots \left( 2 \right)\end{align}\]

And

\[\begin{align}{\left( {1 + \frac{x}{2}} \right)^3} &= {}^3{C_0}{\left( 1 \right)^3} + {}^3{C_1}{\left( 1 \right)^2}\left( {\frac{x}{2}} \right) + {}^3{C_2}\left( 1 \right){\left( {\frac{x}{2}} \right)^2} + {}^3{C_3}{\left( {\frac{x}{2}} \right)^3}\\ &= 1 + \frac{{3x}}{2} + \frac{{3{x^2}}}{4} + \frac{{{x^3}}}{8} \qquad \ldots \left( 3 \right)\end{align}\]

From (1), (2) and (3), we obtain

\[\begin{align}{\left[ {\left( {1 + \frac{x}{2}} \right) - \frac{2}{x}} \right]^4} &= 1 + 2x + \frac{{3{x^2}}}{2} + \frac{{{x^3}}}{2} + \frac{{{x^4}}}{{16}} - \frac{8}{x}\left( {1 + \frac{{3x}}{2} + \frac{{3{x^2}}}{4} + \frac{{{x^3}}}{8}} \right) + \frac{8}{{{x^2}}} + \frac{{24}}{x} + 6 - \frac{{32}}{{{x^3}}} + \frac{{16}}{{{x^4}}}\\&= 1 + 2x + \frac{{3{x^2}}}{2} + \frac{{{x^3}}}{2} + \frac{{{x^4}}}{{16}} - \frac{8}{x} - 12 - 6x - {x^2} + \frac{8}{{{x^2}}} + \frac{{24}}{x} + 6 - \frac{{32}}{{{x^3}}} + \frac{{16}}{{{x^4}}}\\&= \frac{{16}}{x} + \frac{8}{{{x^2}}} - \frac{{32}}{{{x^3}}} + \frac{{16}}{{{x^4}}} - 4x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{2} + \frac{{{x^4}}}{{16}} - 5\end{align}\]

Chapter 8 Ex.8.ME Question 10

Find the expansion of \({\left( {3{x^2} - 2ax + 3{a^2}} \right)^3}\)

 

Solution

Video Solution

 

Using binomial theorem, the expansion of \({\left( {3{x^2} - 2ax + 3{a^2}} \right)^3}\) is given by

\[\begin{align} {{\left[ \left( 3{{x}^{2}}-2ax \right)+3{{a}^{2}} \right]}^{3}}&={}^{3}{{C}_{0}}{{\left( 3{{x}^{2}}-2ax \right)}^{3}}+{}^{3}{{C}_{1}}{{\left( 3{{x}^{2}}-2ax \right)}^{2}}\left( 3{{a}^{2}} \right)\\&\quad+{}^{3}{{C}_{2}}\left( 3{{x}^{2}}-2ax \right){{\left( 3{{a}^{2}} \right)}^{2}}+{}^{3}{{C}_{3}}{{\left( 3{{a}^{2}} \right)}^{3}} \\ & ={{\left( 3{{x}^{2}}-2ax \right)}^{3}}+3\left( 9{{x}^{4}}-12a{{x}^{3}}+4{{a}^{2}}{{x}^{2}} \right)\left( 3{{a}^{2}} \right)\\&\quad+3\left( 3{{x}^{2}}-2ax \right)\left( 9{{a}^{4}} \right)+27{{a}^{6}} \\ & ={{\left( 3{{x}^{2}}-2ax \right)}^{3}}+81{{a}^{2}}{{x}^{4}}-108{{a}^{3}}{{x}^{3}}\\&\quad+36{{a}^{4}}{{x}^{2}}+81{{a}^{4}}{{x}^{2}}-54{{a}^{5}}x+27{{a}^{6}} \\ & ={{\left( 3{{x}^{2}}-2ax \right)}^{3}}+81{{a}^{2}}{{x}^{4}}-108{{a}^{3}}{{x}^{3}}\\&\quad+117{{a}^{4}}{{x}^{2}}-54{{a}^{5}}x+27{{a}^{6}} \qquad \qquad \ldots \left( 1 \right) \end{align}\]

Again, by using binomial theorem, we obtain

\[\begin{align}{\left( {3{x^2} - 2ax} \right)^3} &= {}^3{C_0}{\left( {3{x^2}} \right)^3} - {}^3{C_1}{\left( {3{x^2}} \right)^2}\left( {2ax} \right) + {}^3{C_2}\left( {3{x^2}} \right){\left( {2ax} \right)^2} + {}^3{C_3}{\left( {2ax} \right)^3}\\&= 27{x^6} - 3\left( {9{x^4}} \right)\left( {2ax} \right) + 3\left( {3{x^2}} \right)\left( {4{a^2}{x^2}} \right) - 8{a^3}{x^3}\\&= 27{x^6} - 54a{x^5} + 36{a^2}{x^4} - 8{a^3}{x^3} \qquad \quad \ldots \left( 2 \right)\end{align}\]

From (1) and (2), we obtain

\[\begin{align}{\left( {3{x^2} - 2ax + 3{a^2}} \right)^3} &= 27{x^6} - 54a{x^5} + 36{a^2}{x^4} - 8{a^3}{x^3} + 81{a^2}{x^4} - 108{a^3}{x^3} + 117{a^4}{x^2} - 54{a^5}x + 27{a^6}\\&= 27{x^6} - 54a{x^5} + 117{a^2}{x^4} - 116{a^3}{x^3} + 117{a^4}{x^2} - 54{a^5}x + 27{a^6}\end{align}\]

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