We have (3x² - 2ax + 3a²)³
=[(3x²- 2ax) + 3a²)]³
=³C₀(3x²- 2ax)³ + ³C₁(3x² - 2ax)² (3a²)+³C₂(3x²- 2ax) (3a²)²+ ³C₃(3a²)³
= (3x² - 2ax)³ + 3 x 3a² (3x² - 2ax)² + 3x 9a⁴ (3x² - 2ax) + 27a⁶
= (27x⁶ - 8a³x³ - 54ax⁵ + 36a²x⁴) + 9a² (9x⁴+ 4a²x² - 12ax³) + 27a⁴ (3x² - 2ax) + 27a⁶
= 27x⁶ - 108a³x³- 54ax⁵ + 36a²x⁴ + 81a²x⁴+ 36a⁴x² - 108a³x³ + 81a⁴x² - 54a⁵x + 27a⁶
= 27x⁶ - 54ax⁵ + 117a²x⁴ - 116a³x³
+ 117a⁴x² - 54a⁵x + 27 a⁶

Putting a²=x and \(\sqrt { { a }^{ 2 }-1 } \) =y, we have
\(\left( { a }^{ 2 }+\sqrt { { a }^{ 2 }-1 } \right) ^{ 4 }+({ a }^{ 2 }-\sqrt { { a }^{ 2 }-1)^{ 4 } } \)
=(x+y)⁴+(x-y)⁴
=[⁴C₀x⁴+⁴C₁x³y+⁴C₂x²y²+⁴C₃xy³+⁴C₄y⁴]+[⁴C₀x^4-4C₁x³y+⁴C₂x²y^2-4C₃xy^3-4C₄y⁴]
=2[⁴C₀x⁴+⁴C₂x²y²+⁴C₄y⁴]
=2[x⁴+6x²y²+y⁴]
=2[a²)⁴+6(a²)²\(\left( \sqrt { { a }^{ 2 }-1 } \right) ^{ 2 }\)+\(\left( \sqrt { { a }^{ 2 }-1 } \right) ^{ 4 }\)]
=2[a⁸+6a⁴(a²-1)+(a²-1)²]
=2[a⁸+6a⁶-6a⁴+a⁴-2a²+1]
=2[a⁸+6a⁶-5a⁴-2a²+1]

T_{2 =}T₁₊₁ =ⁿC₁ x^n-1. a=240 .....(i)
Similarly, T₃ =ⁿC₂ x^n-2. a² =720 ......(ii)
and T₄=ⁿC₃ x^n-3 a³ =1080......(iii)
On dividing. Eq. (ii) by Eq.(i), we get
\(\frac { { ^{ n }C }_{ 2 }.x^{ n-2 }.a^{ 2 } }{ { ^{ n }C }_{ 1 }.x^{ n-1 }.a } =\frac { 720 }{ 240 } \Rightarrow \frac { a }{ x } =\frac { 6 }{ n-1 } \)
On dividing Eq.(iii) by Eq. (ii) we get
\(\frac { a }{ x } =\frac { 9 }{ 2(n-2) } \Rightarrow Then,\quad \frac { 6 }{ n-1 } =\frac { 9 }{ 2(n-2) } \Rightarrow n=5\)
Now \(T_{ 2 }={ ^{ n }C }_{ 1 }x^{ 5-1 }.a=240\quad \Rightarrow 5.x^{ 4 }.a=240\)
\(\because \frac { a }{ x } =\frac { 6 }{ n-1 } =\frac { 6 }{ 5-1 } =\frac { 6 }{ 4 } =\frac { 3 }{ 2 }\)
\( \Rightarrow a=\frac { 3 }{ 2 } x\)
\(\therefore 5.x^{ 4 }.\frac { 3 }{ 2 } .x=240\)
Ans. x=2, a=3 and n=5

(1.2)⁴⁰⁰⁰ = (1 +0.2)⁴⁰⁰⁰
= ⁴⁰⁰⁰C₀ +   ⁴⁰⁰⁰C₁(0.2) + sum of positive terms
= 1 + 4000(0.2) + a positive number
= 1 + 800 + positive number
= 800.

99 can be written as the sum or difference of two numbers whose powers are easier to calculate and then, Binomial Theorem can be applied.
It can be written that, 99 = 100 – 1
\(\therefore(99)^{5}=(100-1)^{5} \)
\(={ }^{5} \mathrm{C}_{0}(100)^{5}-{ }^{5} \mathrm{C}_{1}(100)^{4}(1)+{ }^{5} \mathrm{C}_{2}(100)^{3}(1)^{2}-{ }^{5} \mathrm{C}_{3}(100)^{2}(1)^{3} +{ }^{5} \mathrm{C}_{4}(100)(1)^{4}-{ }^{5} \mathrm{C}_{5}(1)^{5} \)
\(=(100)^{5}-5(100)^{4}+10(100)^{3}-10(100)^{2}+5(100)-1 \)
\(= 10000000000-500000000+10000000-100000+500-1\)
\(= 10010000500-500100001 \)
\(=9509900499\)

By using Binomial Theorem, the expression \(\left(x+\frac{1}{x}\right)^{6}\) can be expanded as,
\(\left(x+\frac{1}{x}\right)^{6}={ }^{6} C_{0}(x)^{6}+{ }^{6} C_{1}(x)^{5}\left(\frac{1}{x}\right)+{ }^{6} C_{2}(\mathrm{x})^{4}\left(\frac{1}{x}\right)^{2} \)
\(+{ }^{6} C_{3}(x)^{3}\left(\frac{1}{x}\right)^{3}+{ }^{6} C_{4}(x)^{2}\left(\frac{1}{x}\right)^{4}+{(}_{5}^{6}(x)\left(\frac{1}{x}\right)^{5}+{ }^{6} C_{6}\left(\frac{1}{x}\right)^{6} \)
\(= x^{6}+6(x)^{5}\left(\frac{1}{x}\right)+15(x)^{4}\left(\frac{1}{x^{2}}\right)+20(x)^{3}\left(\frac{1}{x^{3}}\right)+15(x)^{2}\left(\frac{1}{x^{4}}\right)+6(x)\left(\frac{1}{x^{5}}\right)+\frac{1}{x^{6}} \)
\(= x^{6}+6 x^{4}+15 x^{2}+20+\frac{15}{x^{2}}+\frac{6}{x^{4}}+\frac{1}{x^{6}}\)

By using Binomial Theorem, the expression \(\left(\frac{x}{3}+\frac{1}{x}\right)^{5}\) can be expanded as,
\(\left(\frac{x}{3}+\frac{1}{x}\right)^{5} ={ }^{5} C_{0}\left(\frac{x}{3}\right)^{5}+{ }^{5} C_{1}\left(\frac{x}{3}\right)^{4}\left(\frac{1}{x}\right)+{ }^{5} C_{2}\left(\frac{x}{3}\right)^{3}\left(\frac{1}{x}\right)^{2} \)
\(+{ }^{5} C_{3}\left(\frac{x}{3}\right)^{2}\left(\frac{1}{x}\right)^{3}+{ }^{5} C_{4}\left(\frac{x}{3}\right)\left(\frac{1}{x}\right)^{4}+\xi_{5}\left(\frac{1}{x}\right)^{5} \)
\(=\frac{x^{5}}{243}+5\left(\frac{x^{4}}{81}\right)\left(\frac{1}{x}\right)+10\left(\frac{x^{3}}{27}\right)\left(\frac{1}{x^{2}}\right)+10\left(\frac{x^{2}}{9}\right)\left(\frac{1}{x^{3}}\right)+5\left(\frac{x}{3}\right)\left(\frac{1}{x^{4}}\right)+\frac{1}{x^{5}} \)
\(=\frac{x^{5}}{243}+\frac{5 x^{3}}{81}+\frac{10 x}{27}+\frac{10}{9 x}+\frac{5}{3 x^{3}}+\frac{1}{x^{5}}\)

\(=[^{ 6 }{ C }_{ 0 }{ \left( \sqrt { 3 } \right) }^{ 6 }+^{ 6 }{ C }_{ 1 }\left( \sqrt { 3 } \right) ^{ 5 }.\sqrt { 2 } +^{ 6 }C_{ 2 }\left( \sqrt { 3 } \right) ^{ 4 }\left( \sqrt { 2 } \right) ^{ 2 }+^{ 6 }C_{ 3 }\left( \sqrt { 3 } \right) ^{ 3 }\left( \sqrt { 2 } \right) ^{ 3 }+^{ 6 }C_{ 4 }\left( \sqrt { 3 } \right) ^{ 2 }\left( \sqrt { 2 } \right) ^{ 4 }+^{ 6 }C_{ 5 }\left( \sqrt { 3 } \right) \left( \sqrt { 2 } \right) ^{ 5 }+^{ 6 }C_{ 6 }\left( \sqrt { 2 } \right) ^{ 6 }]-[^{ 6 }{ C }_{ 0 }{ \left( \sqrt { 3 } \right) }^{ 6 }\\ -^{ 6 }{ C }_{ 1 }\left( \sqrt { 3 } \right) ^{ 5 }.\sqrt { 2 } +^{ 6 }C_{ 2 }\left( \sqrt { 3 } \right) ^{ 4 }\left( \sqrt { 2 } \right) ^{ 2 }-^{ 6 }C_{ 3 }\left( \sqrt { 3 } \right) ^{ 3 }\left( \sqrt { 2 } \right) ^{ 3 }+^{ 6 }C_{ 4 }\left( \sqrt { 3 } \right) ^{ 2 }\left( \sqrt { 2 } \right) ^{ 4 }-^{ 6 }C_{ 5 }\left( \sqrt { 3 } \right) \left( \sqrt { 2 } \right) ^{ 2 }+^{ 6 }C_{ 6 }\left( \sqrt { 2 } \right) ^{ 6 }]\)
\(=^{ 6 }{ C }_{ 0 }{ \left( \sqrt { 3 } \right) }^{ 6 }+^{ 6 }{ C }_{ 1 }\left( \sqrt { 3 } \right) ^{ 5 }.\sqrt { 2 } +^{ 6 }C_{ 2 }\left( \sqrt { 3 } \right) ^{ 4 }\left( \sqrt { 2 } \right) ^{ 2 }+^{ 6 }C_{ 3 }\left( \sqrt { 3 } \right) ^{ 3 }\left( \sqrt { 2 } \right) ^{ 3 }+^{ 6 }C_{ 4 }\left( \sqrt { 3 } \right) ^{ 2 }\left( \sqrt { 2 } \right) ^{ 4 }+^{ 6 }C_{ 5 }\left( \sqrt { 3 } \right) \left( \sqrt { 2 } \right) ^{ 5 }+^{ 6 }C_{ 6 }\left( \sqrt { 2 } \right) ^{ 6 }-^{ 6 }{ C }_{ 0 }{ \left( \sqrt { 3 } \right) }^{ 6 }\\ +^{ 6 }{ C }_{ 1 }\left( \sqrt { 3 } \right) ^{ 5 }.\sqrt { 2 } -^{ 6 }C_{ 2 }\left( \sqrt { 3 } \right) ^{ 4 }\left( \sqrt { 2 } \right) ^{ 2 }+^{ 6 }C_{ 3 }\left( \sqrt { 3 } \right) ^{ 3 }\left( \sqrt { 2 } \right) ^{ 3 }-^{ 6 }C_{ 4 }\left( \sqrt { 3 } \right) ^{ 2 }\left( \sqrt { 2 } \right) ^{ 4 }+^{ 6 }C_{ 5 }\left( \sqrt { 3 } \right) \left( \sqrt { 2 } \right) ^{ 2 }-^{ 6 }C_{ 6 }\left( \sqrt { 2 } \right) ^{ 6 }\)
\(=2^{ 6 }{ C }_{ 1 }\left( \sqrt { 3 } \right) ^{ 5 }.\sqrt { 2 } +2^{ 6 }{ C }_{ 3 }\left( \sqrt { 3 } \right) ^{ 3 }.\left( \sqrt { 2 } \right) ^{ 3 }+2^{ 6 }{ C }_{ 5 }\left( \sqrt { 3 } \right) .\left( \sqrt { 2 } \right) ^{ 5 }\)
\(=\left( 2\times 6\times 9\sqrt { 3 } \times \sqrt { 2 } \right) +(2\times 20\times 3\sqrt { 3 } \times 2\sqrt { 3 } )+(2\times 6\times \sqrt { 3 } \times 4\sqrt { 2 } )\)
\(=108\sqrt { 6 } +240\sqrt { 6 } +48\sqrt { 6 } =396\sqrt { 6 }\) 

We have 9ⁿ⁺¹ = (1 + 8)ⁿ⁺¹
= ⁿ⁺¹C_o + ⁿ⁺¹C₁(8) + ⁿ⁺¹C₂(8)² + ⁿ⁺¹C₃(8)³ + .... +  ⁿ⁺¹C_n+1(8)ⁿ⁺¹
= 1 + (n + 1) x 8 + ⁿ⁺¹C₂(8)² + ⁿ⁺¹C₃(8)³ + .... + ⁿ⁺¹C_n+1(8)ⁿ⁺¹
= 1+ 8n + 8 +  ⁿ⁺¹C₂(8)² + ⁿ⁺¹C₃(8)³ + .... + ⁿ⁺¹C_n+1(8)ⁿ⁺¹
= 9ⁿ⁺¹ - 8n -9
= ⁿ⁺¹C₂(8)² + ⁿ⁺¹C₃(8)³ + ..... + ⁿ⁺¹C_n+1(8)ⁿ⁺¹
= 64 [ ⁿ⁺¹C₂ +  ⁿ⁺¹C₃ . 8 + ..... + ⁿ⁺¹C_n+1 . 8ⁿ⁺¹]
= which shows that  9ⁿ⁺¹ - 8n - 9 is divisible by 64.

(x + 1)⁶ + (x - 1)⁶ 
= [⁶C_ox⁶ + ⁶C₁x⁵ + ⁶C₂x⁴ + ⁶C₃x³ + ⁶C₄x² + ⁶C₅x + ⁶C₆] + [⁶C_ox⁶ + ⁶C₁x⁵(-1) + ⁶C₂x⁴(-1)² + ⁶C₃x³(-1)³ + ⁶C₄x²(-1)⁴ + ⁶C₅x(-1)⁵+ ⁶C₆(-1)⁶]
= [x⁶ + 6x⁵ + 15x⁴ + 20x³ + 15x² + 6x + 1] - [x⁶ - 6x⁵ + 15x⁴ - 20x³ + 15x² - 6x + 1]
= 2x⁶ + 30x⁴ + 30x² + 2
= 2( x⁶ + 15x⁴ + 15x² + 1)
Putting x = \(\sqrt { 2 } \)
\((\sqrt{2}+1)^{6}+(\sqrt{2}-1)^{6}\)
= 2 [​​​​​(​​\(\sqrt { 2 } \))⁶ + 15\(\left( \sqrt { 2 } \right) ^{ 4 }\) + 15\(\left( \sqrt { 2 } \right) ^{ 2 }\) + 1]
= 2 [8 + 15 x 4 + 15 x 2 + 1]
= 2 [ 8 + 60 + 30 + 1]
= 2 x 99 = 198

For two numbers a and b if we can find numbers q and r such that a = bq + r, then we say that b divides a with q as quotient and r as remainder. Thus, in order to show that 6ⁿ – 5n leaves remainder 1 when divided by 25, we prove that 6ⁿ – 5n = 25k + 1, where k is some natural number
We have
(1 + a)ⁿ = ⁿC₀ + ⁿC₁a + ⁿC₂a² + ... + ⁿC_naⁿ
For a = 5, we get
(1 + 5)ⁿ = ⁿC₀ + ⁿC₁5 + ⁿC₂5² + ... + ⁿC_n5ⁿ
i.e. (6)ⁿ = 1 + 5n + 5².ⁿC₂ + 5³.ⁿC₃ + ... + 5n
i.e. 6ⁿ – 5n = 1+5² (ⁿC₂ + ⁿC₃5 + ... + 5^n-2)
or 6ⁿ – 5n = 1+ 25 (ⁿC₂ + 5 .ⁿC₃ + ... + 5^n-2)
or 6ⁿ – 5n = 25k+1 where k = ⁿC₂ + 5 .ⁿC₃ + ... + 5^n–2.
This shows that when divided by 25, 6ⁿ – 5n leaves remainder 1.

Splitting 1.01 and using binomial theorem to write the first few terms we have
(1.01)^1000000 = (1 + 0.01)^1000000
= ^1000000C₀ + ^1000000C₁(0.01) + other positive terms
= 1 + 1000000 x 0.01 + other positive terms
= 1 + 10000 + other positive terms
> 10000
Hence (1.01)^1000000 > 10000.

We express 98 as the sum or difference of two numbers whose powers are easier to calculate, and then use Binomial Theorem.
Write 98 = 100 – 2
Therefore, (98)⁵ = (100 – 2)⁵
= ⁵C₀ (100)⁵ – ⁵C₁ (100)⁴.2 + ⁵C₂ (100)³2²
– ⁵C₃ (100)² (2)³ + ⁵C₄ (100) (2)⁴ – ⁵C₅ (2)⁵
= 10000000000 – 5 x 100000000 x 2 + 10 x 1000000 x 4 – 10 x10000 x 8 + 5 x 100 x 16 – 32
= 10040008000 – 1000800032 = 9039207968.

By using binomial theorem, we have
\(\left(x^{2}+\frac{3}{x}\right)^{4} ={ }^{4} \mathrm{C}_{0}\left(x^{2}\right)^{4}+{ }^{4} \mathrm{C}_{1}\left(x^{2}\right)^{3}\left(\frac{3}{x}\right)+{ }^{4} \mathrm{C}_{2}\left(x^{2}\right)^{2}\left(\frac{3}{x}\right)^{2}+{ }^{4} \mathrm{C}_{3}\left(x^{2}\right)\left(\frac{3}{x}\right)^{3}+{ }^{4} \mathrm{C}_{4}\left(\frac{3}{x}\right)^{4}\)
\(=x^{8}+4 x^{6} \cdot \frac{3}{x}+6 \cdot x^{4} \cdot \frac{9}{x^{2}}+4 \cdot x^{2} \cdot \frac{27}{x^{3}}+\frac{81}{x^{4}} \)
\(=x^{8}+12 x^{5}+54 x^{2}+\frac{108}{x}+\frac{81}{x^{4}}\)