Answer :

In order to prove that $\left(a-b\right)$ is a factor of $\left(a^{n} - b^{n}\right)$, it has to be proved that $a^{n} - b^{n}=k\left(a - b\right)$,

where $k$ is some natural number It can be written that, $a=a - b+b$

$ \begin{aligned} \therefore \ \ a^{n} & =\left(a-b+b\right)^{n}=\left[\left(a-b\right)+b\right]^{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} \\ \\ & =\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} \\ \\ \Rightarrow \ \ & 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] \\ \\ \Rightarrow \ \ & a^{n}-b^{n}=k\left(a-b\right) \end{aligned} $

where, $k=\left[\left(a-b\right)^{n-1}+{ }^{n} C_1\left(a-b\right)^{n-2} b+\ldots+{ }^{n} C _{n-1} b^{n-1}\right]$ is a natural number

This shows that $\left(a-b \right)$ is a factor of $\left(a^{n} - b^{n}\right)$, where $n$ is a positive integer.

Answer :

Firstly, the expression $\left(a+b\right)^{6} -\left(a - b\right)^{6}$ is simplified by using Binomial Theorem.

This can be done as

$ \begin{aligned} \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^{1} 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}+6 a 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^{1} 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}-6 a b^{5}+b^{6} \\ \\ \therefore\quad & \left(a+b\right)^{6}-\left(a-b\right)^{6} =2\left[6 a^{5} b+20 a^{3} b^{3}+6 a b^{5}\right] \end{aligned} $

Putting $a=\sqrt{3}$ and $b=\sqrt{2}$, we obtain

$ \begin{aligned} \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{aligned} $

Answer :

Firstly, the expression $\left(x+y\right)^{4}+\left(x - y\right)^{4}$ is simplified by using Binomial Theorem.

This can be done as

$ \begin{aligned} \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^{3}+{ }^{4} C_4 y^{4} \\ \\ & =x^{4}+4 x^{3} y+6 x^{2} y^{2}+4 x 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^{3}+{ }^{4} C_4 y^{4} \\ \\ & =x^{4}-4 x^{3} y+6 x^{2} y^{2}-4 x y^{3}+y^{4} \\ \\ \therefore \quad & \left(x+y\right)^{4}+\left(x-y\right)^{4}=2\left(x^{4}+6 x^{2} y^{2}+y^{4}\right) \end{aligned} $

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

$ \begin{aligned} \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{aligned} $

Answer :

$0.99=1-0.01$

$ \begin{aligned} \therefore \ \ \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{aligned} $

(Approximately)

Thus, the value of $\left(0.99\right)^{5}$ is approximately $0.951 .$

Answer :

Using Binomial Theorem, the given expression $\left(1+\dfrac{x}{2}-\dfrac{2}{x}\right)^{4}$ can be expanded as

$ \begin{aligned} & {\left[\left(1+\dfrac{x}{2}\right)-\dfrac{2}{x}\right]^{4}} ={ }^{4} C_0\left(1+\dfrac{x}{2}\right)^{4}-{ }^{4} C_1\left(1+\dfrac{x}{2}\right)^{3}\left(\dfrac{2}{x}\right)+{ }^{4} C_2\left(1+\dfrac{x}{2}\right)^{2}\left(\dfrac{2}{x}\right)^{2}-{ }^{4} C_3\left(1+\dfrac{x}{2}\right)\left(\dfrac{2}{x}\right)^{3}+{ }^{4} C_4\left(\dfrac{2}{x}\right)^{4} \\ \\ & \hspace{2.6cm} =\left(1+\dfrac{x}{2}\right)^{4}-4\left(1+\dfrac{x}{2}\right)^{3}\left(\dfrac{2}{x}\right)+6\left(1+x+\dfrac{x^{2}}{4}\right)\left(\dfrac{4}{x^{2}}\right)-4\left(1+\dfrac{x}{2}\right)\left(\dfrac{8}{x^{3}}\right)+\dfrac{16}{x^{4}} \\ \\ & \hspace{2.6cm} =\left(1+\dfrac{x}{2}\right)^{4}-\dfrac{8}{x}\left(1+\dfrac{x}{2}\right)^{3}+\dfrac{24}{x^{2}}+\dfrac{24}{x}+6-\dfrac{32}{x^{3}}-\dfrac{16}{x^{2}}+\dfrac{16}{x^{4}} \\ \\ & \hspace{2.6cm} =\left(1+\dfrac{x}{2}\right)^{4}-\dfrac{8}{x}\left(1+\dfrac{x}{2}\right)^{3}+\dfrac{8}{x^{2}}+\dfrac{24}{x}+6-\dfrac{32}{x^{3}}+\dfrac{16}{x^{4}} \qquad \ldots {\left(1\right)} \end{aligned} $

Again by using Binomial Theorem, we obtain

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

From $\left(1\right), \ \left(2\right),$ and $\left(3\right),$ we obtain

$ \begin{aligned} & {\left[\left(1+\dfrac{x}{2}\right)-\dfrac{2}{x}\right]^{4}} =1+2 x+\dfrac{3 x^{2}}{2}+\dfrac{x^{3}}{2}+\dfrac{x^{4}}{16}-\dfrac{8}{x}\left(1+\dfrac{3 x}{2}+\dfrac{3 x^{2}}{4}+\dfrac{x^{3}}{8}\right)+\dfrac{8}{x^{2}}+\dfrac{24}{x}+6-\dfrac{32}{x^{3}}+\dfrac{16}{x^{4}} \\ \\ &\hspace{2.6cm} =1+2 x+\dfrac{3}{2} x^{2}+\dfrac{x^{3}}{2}+\dfrac{x^{4}}{16}-\dfrac{8}{x}-12-6 x-x^{2}+\dfrac{8}{x^{2}}+\dfrac{24}{x}+6-\dfrac{32}{x^{3}}+\dfrac{16}{x^{4}} \\ \\ &\hspace{2.6cm} =\dfrac{16}{x}+\dfrac{8}{x^{2}}-\dfrac{32}{x^{3}}+\dfrac{16}{x^{4}}-4 x+\dfrac{x^{2}}{2}+\dfrac{x^{3}}{2}+\dfrac{x^{4}}{16}-5 \end{aligned} $

Answer :

Using Binomial Theorem, the given expression $\left(3 x^{2}-2 a x+3 a^{2}\right)^{3}$ can be expanded as

$\left[\left(3 x^{2}-2 ax\right)+3 a^{2}\right]^{3}$ $={ }^{3} C_0\left(3 x^{2}-2 ax\right)^{3}+{ }^{3} C_1\left(3 x^{2}-2 ax\right)^{2}\left(3 a^{2}\right)+{ }^{3} C_2\left(3 x^{2}-2 ax\right)\left(3 a^{2}\right)^{2}+{ }^{3} C_3\left(3 a^{2}\right)^{3}$

$\hspace{3.2cm} =\left(3 x^{2}-2 ax\right)^{3}+3\left(9 x^{4}-12 ax^{3}+4 a^{2} x^{2}\right)\left(3 a^{2}\right)+3\left(3 x^{2}-2 ax\right)\left(9 a^{4}\right)+27 a^{6}$

$\hspace{3.2cm} =\left(3 x^{2}-2 ax\right)^{3}+81 a^{2} x^{4}-108 a^{3} x^{3}+36 a^{4} x^{2}+81 a^{4} x^{2}-54 a^{5} x+27 a^{6}$

$\hspace{3.2cm} =\left(3 x^{2}-2 ax\right)^{3}+81 a^{2} x^{4}-108 a^{3} x^{3}+117 a^{4} x^{2}-54 a^{5} x+27 a^{6}\qquad \ldots(1)$

Again by using Binomial Theorem, we obtain

$ \begin{aligned} & \left(3 x^{2}-2 a x\right)^{3} ={ }^{3} C_0\left(3 x^{2}\right)^{3}-{ }^{3} C_1\left(3 x^{2}\right)^{2}\left(2 a x\right)+{ }^{3} C_2\left(3 x^{2}\right)\left(2 a x\right)^{2}-{ }^{3} C_3\left(2 a x\right)^{3} \\ \\ & \hspace{2cm} =27 x^{6}-3\left(9 x^{4}\right)\left(2 a x\right)+3\left(3 x^{2}\right)\left(4 a^{2} x^{2}\right)-8 a^{3} x^{3} \\ \\ & \hspace{2cm} =27 x^{6}-54 a x^{5}+36 a^{2} x^{4}-8 a^{3} x^{3} \qquad \ldots {\left(2\right)} \end{aligned} $

From $\left(1\right)$ and $\left(2\right),$ we obtain

$ \begin{aligned} & \left(3 x^{2}-2 a x+3 a^{2}\right)^{3} =27 x^{6}-54 a 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} \\ \\ & \hspace{2.9cm} =27 x^{6}-54 a 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{aligned} $