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The concepts covered in Chapter 8 of the Maths textbook includes the study of essential topics such as Positive Integral Indices, Pascal’s Triangle, Binomial theorem for any positive integer and some special cases. Students can score high marks in the exams with ease by practising the NCERT Solutions for all the questions present in the textbook. Each solution is solved step-by-step, considering the understanding level of the students. Therefore, it is important to understand the logic set behind each answer and develop a better comprehension of the concepts.

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Exercise 8.1 Page No: 166

**Expand each of the expressions in Exercises 1 to 5.**

**1. (1 – 2x) ^{5}**

**Solution:**

From binomial theorem expansion we can write as

(1 – 2x)^{5}

= ^{5}C_{o }(1)^{5} – ^{5}C_{1 }(1)^{4} (2x) + ^{5}C_{2 }(1)^{3 }(2x)^{2} – ^{5}C_{3 }(1)^{2 }(2x)^{3} + ^{5}C_{4 }(1)^{1} (2x)^{4} – ^{5}C_{5 }(2x)^{5}

= 1 – 5 (2x) + 10 (4x)^{2} – 10 (8x^{3}) + 5 ( 16 x^{4}) – (32 x^{5})

= 1 – 10x + 40x^{2} – 80x^{3} + 80x^{4}– 32x^{5}

**Solution:**

From binomial theorem, given equation can be expanded as

**3. (2x – 3) ^{6}**

**Solution:**

From binomial theorem, given equation can be expanded as

**Solution:**

From binomial theorem, given equation can be expanded as

**Solution:**

From binomial theorem, given equation can be expanded as

**6. (96) ^{3}**

**Solution:**

Given (96)^{3}

96 can be expressed as the sum or difference of two numbers and then binomial theorem can be applied.

The given question can be written as 96 = 100 – 4

(96)^{3} = (100 – 4)^{3}

= ^{3}C_{0} (100)^{3} – ^{3}C_{1} (100)^{2} (4) – ^{3}C_{2} (100) (4)^{2}– ^{3}C_{3} (4)^{3}

= (100)^{3} – 3 (100)^{2} (4) + 3 (100) (4)^{2} – (4)^{3}

= 1000000 – 120000 + 4800 – 64

= 884736

**7. (102) ^{5}**

**Solution:**

Given (102)^{5}

102 can be expressed as the sum or difference of two numbers and then binomial theorem can be applied.

The given question can be written as 102 = 100 + 2

(102)^{5} = (100 + 2)^{5}

= ^{5}C_{0} (100)^{5} + ^{5}C_{1} (100)^{4} (2) + ^{5}C_{2} (100)^{3} (2)^{2} + ^{5}C_{3} (100)^{2} (2)^{3} + ^{5}C_{4} (100) (2)^{4} + ^{5}C_{5} (2)^{5}

= (100)^{5} + 5 (100)^{4} (2) + 10 (100)^{3} (2)^{2} + 5 (100) (2)^{3} + 5 (100) (2)^{4} + (2)^{5}

= 1000000000 + 1000000000 + 40000000 + 80000 + 8000 + 32

= 11040808032

**8. (101) ^{4}**

**Solution:**

Given (101)^{4}

101 can be expressed as the sum or difference of two numbers and then binomial theorem can be applied.

The given question can be written as 101 = 100 + 1

(101)^{4} = (100 + 1)^{4}

= ^{4}C_{0} (100)^{4} + ^{4}C_{1} (100)^{3} (1) + ^{4}C_{2} (100)^{2} (1)^{2} + ^{4}C_{3} (100) (1)^{2} + ^{4}C_{4 }(1)^{4}

= (100)^{4} + 4 (100)^{3} + 6 (100)^{2} + 4 (100) + (1)^{4}

= 100000000 + 400000 + 60000 + 400 + 1

= 1040604001

**9. (99) ^{5}**

**Solution:**

Given (99)^{5}

99 can be written as the sum or difference of two numbers then binomial theorem can be applied.

The given question can be written as 99 = 100 -1

(99)^{5} = (100 – 1)^{5}

= ^{5}C_{0} (100)^{5} – ^{5}C_{1} (100)^{4} (1) + ^{5}C_{2} (100)^{3} (1)^{2} – ^{5}C_{3} (100)^{2} (1)^{3} + ^{5}C_{4} (100) (1)^{4} – ^{5}C_{5} (1)^{5}

= (100)^{5} – 5 (100)^{4} + 10 (100)^{3} – 10 (100)^{2} + 5 (100) – 1

= 1000000000 – 5000000000 + 10000000 – 100000 + 500 – 1

= 9509900499

**10. Using Binomial Theorem, indicate which number is larger (1.1) ^{10000} or 1000.**

**Solution:**

By splitting the given 1.1 and then applying binomial theorem, the first few terms of (1.1)^{10000} can be obtained as

(1.1)^{10000} = (1 + 0.1)^{10000}

= (1 + 0.1)^{10000} C_{1 }(1.1) + other positive terms

= 1 + 10000 × 1.1 + other positive terms

= 1 + 11000 + other positive terms

> 1000

(1.1)^{10000} > 1000

**11.** **Find (a + b) ^{4} – (a – b)^{4}. Hence, evaluate **

**Solution:**

Using binomial theorem the expression (a + b)^{4} and (a – b)^{4}, can be expanded

(a + b)^{4} = ^{4}C_{0} a^{4} + ^{4}C_{1} a^{3} b + ^{4}C_{2} a^{2} b^{2} + ^{4}C_{3} a b^{3} + ^{4}C_{4} b^{4}

(a – b)^{4 }= ^{4}C_{0} a^{4} – ^{4}C_{1} a^{3} b + ^{4}C_{2} a^{2} b^{2} – ^{4}C_{3} a b^{3} + ^{4}C_{4} b^{4}

Now (a + b)^{4} – (a – b)^{4} = ^{4}C_{0} a^{4} + ^{4}C_{1} a^{3} b + ^{4}C_{2} a^{2} b^{2} + ^{4}C_{3} a b^{3} + ^{4}C_{4} b^{4} – [^{4}C_{0} a^{4} – ^{4}C_{1} a^{3} b + ^{4}C_{2} a^{2} b^{2} – ^{4}C_{3} a b^{3} + ^{4}C_{4} b^{4}]

= 2 (^{4}C_{1} a^{3} b + ^{4}C_{3} a b^{3})

= 2 (4a^{3} b + 4ab^{3})

= 8ab (a^{2} + b^{2})

Now by substituting a = √3 and b = √2 we get

(√3 + √2)^{4} – (√3 – √2)^{4} = 8 (√3) (√2) {(√3)^{2} + (√2)^{2}}

= 8 (√6) (3 + 2)

= 40 √6

**12.** **Find (x + 1) ^{6} + (x – 1)^{6}. Hence or otherwise evaluate **

** **

**Solution:**

Using binomial theorem the expressions, (x + 1)^{6} and (x – 1)^{6} can be expressed as

(x + 1)^{6} = ^{6}C_{0} x^{6} + ^{6}C_{1} x^{5} + ^{6}C_{2} x^{4} + ^{6}C_{3} x^{3} + ^{6}C_{4} x^{2} + ^{6}C_{5} x + ^{6}C_{6}

(x – 1)^{6} = ^{6}C_{0} x^{6} – ^{6}C_{1} x^{5} + ^{6}C_{2} x^{4} – ^{6}C_{3} x^{3} + ^{6}C_{4} x^{2} – ^{6}C_{5} x + ^{6}C_{6}

Now, (x + 1)^{6} – (x – 1)^{6} = ^{6}C_{0} x^{6} + ^{6}C_{1} x^{5} + ^{6}C_{2} x^{4} + ^{6}C_{3} x^{3} + ^{6}C_{4} x^{2} + ^{6}C_{5} x + ^{6}C_{6} – [^{6}C_{0} x^{6} – ^{6}C_{1} x^{5} + ^{6}C_{2} x^{4} – ^{6}C_{3} x^{3} + ^{6}C_{4} x^{2} – ^{6}C_{5} x + ^{6}C_{6}]

= 2 [^{6}C_{0} x^{6 }+ ^{6}C_{2} x^{4} + ^{6}C_{4} x^{2} + ^{6}C_{6}]

= 2 [x^{6} + 15x^{4} + 15x^{2} + 1]

Now by substituting x = √2 we get

(√2 + 1)^{6} – (√2 – 1)^{6} = 2 [(√2)^{6} + 15(√2)^{4} + 15(√2)^{2} + 1]

= 2 (8 + 15 × 4 + 15 × 2 + 1)

= 2 (8 + 60 + 30 + 1)

= 2 (99)

= 198

**13. Show that 9 ^{n+1} – 8n – 9 is divisible by 64, whenever n is a positive integer.**

**Solution:**

In order to show that 9^{n+1} – 8n – 9 is divisible by 64, it has to be show that 9^{n+1} – 8n – 9 = 64 k, where k is some natural number

Using binomial theorem,

(1 + a)^{m} = ^{m}C_{0} + ^{m}C_{1} a + ^{m}C_{2} a^{2} + …. + ^{m }C _{m} a^{m}

For a = 8 and m = n + 1 we get

(1 + 8)^{n+1} = ^{n+1}C_{0} + ^{n+1}C_{1} (8) + ^{n+1}C_{2} (8)^{2} + …. + ^{n+1 }C _{n+1 }(8)^{n+1}

9^{n+1} = 1 + (n + 1) 8 + 8^{2} [^{n+1}C_{2} + ^{n+1}C_{3} (8) + …. + ^{n+1 }C _{n+1 }(8)^{n-1}]

9^{n+1} = 9 + 8n + 64 [^{n+1}C_{2} + ^{n+1}C_{3} (8) + …. + ^{n+1 }C _{n+1 }(8)^{n-1}]

9^{n+1} – 8n – 9 = 64 k

Where k = [^{n+1}C_{2} + ^{n+1}C_{3} (8) + …. + ^{n+1 }C _{n+1 }(8)^{n-1}] is a natural number

Thus, 9^{n+1} – 8n – 9 is divisible by 64, whenever n is positive integer.

Hence the proof

**14. Prove that **

**Solution:**

Exercise 8.2 Page No: 171

**Find the coefficient of**

**1. x ^{5} in (x + 3)^{8}**

**Solution:**

The general term T_{r+1} in the binomial expansion is given by T_{r+1} = ^{n }C _{r} a^{n-r} b^{r}

Here x^{5} is the T_{r+1} term so a= x, b = 3 and n =8

T_{r+1} = ^{8}C_{r} x^{8-r} 3^{r}…………… (i)

For finding out x^{5}

We have to equate x^{5}= x^{8-r}

⇒ r= 3

Putting value of r in (i) we get

= 1512 x^{5}

Hence the coefficient of x^{5}= 1512

**2. a ^{5}b^{7} in (a – 2b)^{12} .**

**Solution:**

The general term T_{r+1} in the binomial expansion is given by T_{r+1} = ^{n }C _{r} a^{n-r} b^{r}

Here a = a, b = -2b & n =12

Substituting the values, we get

T_{r+1} = ^{12}C_{r} a^{12-r} (-2b)^{r}………. (i)

To find a^{5}

We equate a^{12-r} =a^{5}

r = 7

Putting r = 7 in (i)

T_{8} = ^{12}C_{7} a^{5} (-2b)^{7}

= -101376 a^{5} b^{7}

Hence the coefficient of a^{5}b^{7}= -101376

**Write the general term in the expansion of**

**3. (x ^{2} – y)^{6}**

**Solution:**

The general term T_{r+1} in the binomial expansion is given by

T_{r+1} = ^{n }C _{r} a^{n-r} b^{r}…….. (i)

Here a = x^{2} , n = 6 and b = -y

Putting values in (i)

T_{r+1} = ^{6}C_{r} x ^{2(6-r)} (-1)^{r} y^{r}

= -1^{r 6}c_{r }.x^{12 – 2r}. y^{r}

**4.** **(x ^{2} – y x)^{12}, x ≠ 0.**

**Solution:**

The general term T_{r+1} in the binomial expansion is given by T_{r+1} = ^{n }C _{r} a^{n-r} b^{r}

Here n = 12, a= x^{2} and b = -y x

Substituting the values we get

T_{n+1} =^{12}C_{r} × x^{2(12-r)} (-1)^{r} y^{r} x^{r}

= -1^{r 12}c_{r }.x^{24 –2r}. y^{r}

**5. Find the 4th term in the expansion of (x – 2y) ^{12}.**

**Solution:**

_{r+1} in the binomial expansion is given by T_{r+1} = ^{n }C _{r} a^{n-r} b^{r}

Here a= x, n =12, r= 3 and b = -2y

By substituting the values we get

T_{4} = ^{12}C_{3} x^{9} (-2y)^{3}

= -1760 x^{9} y^{3}

**6.** **Find the 13 ^{th} term in the expansion of **

**Solution:**

**Find the middle terms in the expansions of**

**Solution:**

**Solution:**

**9. In the expansion of (1 + a) ^{m+n}, prove that coefficients of a^{m} and a^{n} are equal.**

**Solution:**

We know that the general term T_{r+1} in the binomial expansion is given by T_{r+1} = ^{n}C_{r} a^{n-r} b^{r}

Here n= m+n, a = 1 and b= a

Substituting the values in the general form

T_{r+1} = ^{m+n }C_{r} 1^{m+n-r} a^{r}

= ^{m+n }C_{r} a^{r}…………. (i)

Now we have that the general term for the expression is,

T_{r+1} = ^{m+n }C_{r} a^{r}

Now, For coefficient of a^{m}

T_{m+1} = ^{m+n }C_{m} a^{m}

Hence, for coefficient of a^{m}, value of r = m

So, the coefficient is ^{m+n }C _{m}

Similarly, Coefficient of a^{n} is ^{m+n }C _{n}

**10. The coefficients of the (r – 1) ^{th}, r^{th} and (r + 1)^{th} terms in the expansion of (x + 1)^{n} are in the ratio 1 : 3 : 5. Find n and r.**

**Solution:**

The general term T_{r+1} in the binomial expansion is given by T_{r+1} = ^{n}C_{r} a^{n-r} b^{r}

Here the binomial is (1+x)^{n} with a = 1 , b = x and n = n

The (r+1)^{th} term is given by

T_{(r+1)} = ^{n}C_{r} 1^{n-r} x^{r}

T_{(r+1)} = ^{n}C_{r} x^{r}

The coefficient of (r+1)^{th} term is ^{n}C_{r}

The r^{th} term is given by (r-1)^{th} term

T_{(r+1-1)} = ^{n}C_{r-1} x^{r-1}

T_{r} = ^{n}C_{r-1} x^{r-1}

∴ the coefficient of r^{th} term is ^{n}C_{r-1}

For (r-1)^{th} term we will take (r-2)^{th} term

T_{r-2+1} = ^{n}C_{r-2} x^{r-2}

T_{r-1} = ^{n}C_{r-2} x^{r-2}

∴ the coefficient of (r-1)^{th} term is ^{n}C_{r-2}

Given that the coefficient of (r-1)^{th}, r^{th} and r+1^{th} term are in ratio 1:3:5

Therefore,

⇒ 5r = 3n – 3r + 3

⇒ 8r – 3n – 3 =0………….2

We have 1 and 2 as

n – 4r ± 5 =0…………1

8r – 3n – 3 =0…………….2

Multiplying equation 1 by number 2

2n -8r +10 =0……………….3

Adding equation 2 and 3

2n -8r +10 =0

-3n – 8r – 3 =0

⇒ -n = -7

n =7 and r = 3

**11. Prove that the coefficient of x ^{n} in the expansion of (1 + x)^{2n} is twice the coefficient of x^{n} in the expansion of (1 + x)^{2n – 1}.**

**Solution:**

The general term T_{r+1} in the binomial expansion is given by T_{r+1} = ^{n}C_{r} a^{n-r} b^{r}

The general term for binomial (1+x)^{2n} is

T_{r+1} = ^{2n}C_{r} x^{r} …………………..1

To find the coefficient of x^{n}

r = n

T_{n+1} = ^{2n}C_{n} x^{n}

The coefficient of x^{n} = ^{2n}C_{n}

The general term for binomial (1+x)^{2n-1} is

T_{r+1} = ^{2n-1}C_{r} x^{r}

To find the coefficient of x^{n}

Putting n = r

T_{r+1} = ^{2n-1}C_{r} x^{n}

The coefficient of x^{n} = ^{2n-1}C_{n}

We have to prove

Coefficient of x^{n} in (1+x)^{2n} = 2 coefficient of x^{n} in (1+x)^{2n-1}

Consider LHS = ^{2n}C_{n}

**12. Find a positive value of m for which the coefficient of x ^{2} in the expansion (1 + x)^{m} is 6.**

**Solution:**

The general term T_{r+1} in the binomial expansion is given by T_{r+1} = ^{n}C_{r} a^{n-r} b^{r}

Here a = 1, b = x and n = m

Putting the value

T_{r+1} = ^{m }C_{r} 1^{m-r} x^{r}

= ^{m }C_{r} x^{r}

We need coefficient of x^{2}

∴ putting r = 2

T_{2+1} = ^{m}C_{2} x^{2}

The coefficient of x^{2} = ^{m}C_{2}

Given that coefficient of x^{2} = ^{m}C_{2} = 6

⇒ m (m – 1) = 12

⇒ m^{2}– m – 12 =0

⇒ m^{2}– 4m + 3m – 12 =0

⇒ m (m – 4) + 3 (m – 4) = 0

⇒ (m+3) (m – 4) = 0

⇒ m = – 3, 4

We need positive value of m so m = 4

Miscellaneous Exercise Page No: 175

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

**Solution:**

We know that (r + 1)^{th} term, (T_{r+1}), in the binomial expansion of (a + b)^{n} is given by

T_{r+1} = ^{n}C_{r} a^{n-t} b^{r}

The first three terms of the expansion are given as 729, 7290 and 30375 respectively. Then we have,

T_{1} = ^{n}C_{0} a^{n-0} b^{0} = a^{n} = 729….. 1

T_{2} = ^{n}C_{1} a^{n-1} b^{1 }= na^{n-1} b = 7290…. 2

T_{3} = ^{n}C_{2} a^{n-2} b^{2} = {n (n -1)/2 }a^{n-2} b^{2} = 30375……3

Dividing 2 by 1 we get

Dividing 3 by 2 we get

From 4 and 5 we have

n. 5/3 = 10

n = 6

Substituting n = 6 in 1 we get

a^{6} = 729

a = 3

From 5 we have, b/3 = 5/3

b = 5

Thus a = 3, b = 5 and n = 76

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

**Solution:**

**3. Find the coefficient of x ^{5} in the product (1 + 2x)^{6} (1 – x)^{7} using binomial theorem.**

**Solution:**

(1 + 2x)^{6} = ^{6}C_{0 }+ ^{6}C_{1} (2x) + ^{6}C_{2} (2x)^{2} + ^{6}C_{3} (2x)^{3} + ^{6}C_{4} (2x)^{4} + ^{6}C_{5} (2x)^{5} + ^{6}C_{6} (2x)^{6}

= 1 + 6 (2x) + 15 (2x)^{2} + 20 (2x)^{3} + 15 (2x)^{4} + 6 (2x)^{5} + (2x)^{6}

= 1 + 12 x + 60x^{2} + 160 x^{3} + 240 x^{4} + 192 x^{5} + 64x^{6}

(1 – x)^{7} = ^{7}C_{0} – ^{7}C_{1} (x) + ^{7}C_{2 }(x)^{2} – ^{7}C_{3 }(x)^{3} + ^{7}C_{4 }(x)^{4} – ^{7}C_{5 }(x)^{5} + ^{7}C_{6 }(x)^{6 }– ^{7}C_{7 }(x)^{7}

= 1 – 7x + 21x^{2} – 35x^{3} + 35x^{4} – 21x^{5} + 7x^{6} – x^{7}

(1 + 2x)^{6} (1 – x)^{7} = (1 + 12 x + 60x^{2} + 160 x^{3} + 240 x^{4} + 192 x^{5} + 64x^{6}) (1 – 7x + 21x^{2} – 35x^{3} + 35x^{4} – 21x^{5} + 7x^{6} – x^{7})

192 – 21 = 171

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

**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} = (a – b + b)^{n} and expand]**

**Solution:**

In order to prove that (a – b) is a factor of (a^{n} – b^{n}), it has to be proved that

a^{n} – b^{n} = k (a – b) where k is some natural number.

a can be written as a = a – b + b

a^{n} = (a – b + b)^{n } = [(a – b) + b]^{n}

= ^{n}C_{0} (a – b)^{n} + ^{n}C_{1} (a – b)^{n-1 } b + …… + ^{n }C _{n} b^{n}

a^{n} – b^{n} = (a – b) [(a –b)^{n-1} + ^{n}C_{1} (a – b)^{n-1 } b + …… + ^{n }C _{n} b^{n}]

a^{n} – b^{n} = (a – b) k

Where k = [(a –b)^{n-1} + ^{n}C_{1} (a – b)^{n-1 } b + …… + ^{n }C _{n} b^{n}] is a natural number

This shows that (a – b) is a factor of (a^{n} – b^{n}), where n is positive integer.

**5. Evaluate **

**Solution:**

Using binomial theorem the expression (a + b)^{6} and (a – b)^{6}, can be expanded

(a + b)^{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 – b)^{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}

Now (a + b)^{6} – (a – b)^{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} – [^{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}]

Now by substituting a = √3 and b = √2 we get

(√3 + √2)^{6} – (√3 – √2)^{6} = 2 [6 (√3)^{5} (√2) + 20 (√3)^{3} (√2)^{3} + 6 (√3) (√2)^{5}]

= 2 [54(√6) + 120 (√6) + 24 √6]

= 2 (√6) (198)

= 396 √6

**6. Find the value of **

**Solution:**

**7. Find an approximation of (0.99) ^{5} using the first three terms of its expansion.**

**Solution:**

0.99 can be written as

0.99 = 1 – 0.01

Now by applying binomial theorem we get

(o. 99)^{5} = (1 – 0.01)^{5}

= ^{5}C_{0 }(1)^{5} – ^{5}C_{1 }(1)^{4} (0.01) + ^{5}C_{2 }(1)^{3} (0.01)^{2 }

= 1 – 5 (0.01) + 10 (0.01)^{2}

= 1 – 0.05 + 0.001

= 0.951

**8. Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of is √6: 1**

**Solution:**

**9. Expand using Binomial Theorem **

**Solution:**

Using binomial theorem the given expression can be expanded as

Again by using binomial theorem to expand the above terms we get

From equation 1, 2 and 3 we get

**10. Find the expansion of (3x ^{2} – 2ax + 3a^{2})^{3} using binomial theorem.**

**Solution:**

We know that (a + b)^{3} = a^{3} + 3a^{2}b + 3ab^{2} + b^{3}

Putting a = 3x^{2} & b = -a (2x-3a), we get

^{2}+ (-a (2x-3a))]

^{3}

= (3x^{2})^{3}+3(3x^{2})^{2}(-a (2x-3a)) + 3(3x^{2}) (-a (2x-3a))^{2} + (-a (2x-3a))^{3}

= 27x^{6} – 27ax^{4 }(2x-3a) + 9a^{2}x^{2 }(2x-3a)^{2} – a^{3}(2x-3a)^{3}

= 27x^{6} – 54ax^{5} + 81a^{2}x^{4} + 9a^{2}x^{2 }(4x^{2}-12ax+9a^{2}) – a^{3 }[(2x)^{3} – (3a)^{3} – 3(2x)^{2}(3a) + 3(2x)(3a)^{2}]

= 27x^{6} – 54ax^{5} + 81a^{2}x^{4} + 36a^{2}x^{4} – 108a^{3}x^{3} + 81a^{4}x^{2} – 8a^{3}x^{3} + 27a^{6} + 36a^{4}x^{2} – 54a^{5}x

= 27x^{6} – 54ax^{5}+ 117a^{2}x^{4} – 116a^{3}x^{3} + 117a^{4}x^{2} – 54a^{5}x + 27a^{6}

Thus, (3x^{2} – 2ax + 3a^{2})^{3}

= 27x^{6} – 54ax^{5}+ 117a^{2}x^{4} – 116a^{3}x^{3} + 117a^{4}x^{2} – 54a^{5}x + 27a^{6}

## NCERT Solutions for Class 11 Maths Chapter 8- Binomial Theorem

The Chapter 8 Binomial Theorem of NCERT Solutions for Class 11 covers the topics given below.

8.1 Introduction to Binomial Theorem

8.2 Binomial Theorem for Positive Integral Indices

Pascal’s Triangle

8.2.1 Binomial theorem for any positive integer n,

8.2.2 Some special cases

8.3 General and Middle Terms

Exercise 8.1 Solutions 14 Questions

Exercise 8.2 Solutions 12 Questions

Miscellaneous Exercise On Chapter 8 Solutions 10 Questions

## NCERT Solutions for Class 11 Maths Chapter 8- Binomial Theorem

The unit Algebra houses the chapter Binomial Theorem, adding up to 30 marks of the total 80 marks. A total of 3 exercises including the miscellaneous exercise is present in this chapter. Chapter 8 of NCERT Solutions for Class 11 Maths discusses the concepts provided underneath:

- The expansion of a binomial for any positive integral n is given by the Binomial Theorem, which is (a+b)
^{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}. - The coefficients of the expansions are arranged in an array. This array is called Pascal’s triangle.
- The general term of an expansion (a + b)
^{n}is T_{r + 1}=^{n}C_{r}a^{n – r}. b^{r}

Therefore, it is ensured that a student who is thorough with the eighth chapter of Class 11, the Binomial Theorem, will be well versed in the history of Binomial Theorem, statement and proof of the binomial theorem for positive integral indices, Pascal’s triangle, General and middle term in binomial expansion as well as simple applications of Binomial theorem.