Question

:) (1+1)^3 - (1 - 1)^3​

Answer 1

(1+1)^3 - (1 - 1)^3

= 2^3 - (0)^3

= 8 - 0

= 8 (ans.)

Answer 2

$\large\underline{\sf{Solution-}}$

Given

$\red{\rm :\longmapsto\: {(1 + i)}^{3} - {(1 - i)}^{3}}$

Let Consider,

$\rm :\longmapsto\: {(1 + i)}^{3}$

We know that,

$\boxed{ \bf{ \: {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}}$

So, using this identity, we get

$\rm \:  =  \:  \: {1}^{3} + {i}^{3} + 3 \times 1 \times i(1 + i)$

We know,

$\boxed{ \bf{ \: {i}^{3} = - \: i}}$

$\rm \:  =  \:  \: 1 - i + 3i + 3 {i}^{2}$

We know,

$\boxed{ \bf{ \: {i}^{2} = - \: 1}}$

So, using this, we get

$\rm \:  =  \:  \: 1 + 2i - 3$

$\rm \:  =  \:  \: - 2 + 2i$

$\bf\implies \: {(1 + i)}^{3} = - 2 + 2i - - - (1)$

Now, Consider,

$\rm :\longmapsto\: {(1 - i)}^{3}$

We know that

$\boxed{ \bf{ \: {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y)}}$

So, using this identity, we get

$\rm \:  =  \:  \: {1}^{3} - {i}^{3} - 3 \times 1 \times i(1 - i)$

We know,

$\boxed{ \bf{ \: {i}^{3} = - \: i}}$

So, using this, we get

$\rm \:  =  \:  \: 1 + i - 3i + 3 {i}^{2}$

We know,

$\boxed{ \bf{ \: {i}^{2} = - \: 1}}$

So, using this we get

$\rm \:  =  \:  \: 1 - 2i - 3$

$\rm \:  =  \:  \: - 2 - 2i$

$\bf\implies \: {(1 - i)}^{3} = - 2 - 2i - - - (2)$

So, Now consider,

$\red{\rm :\longmapsto\: {(1 + i)}^{3} - {(1 - i)}^{3}}$

So, on substituting the values, we get

$\rm \:  =  \:  \: - 2 + 2i - ( - 2 - 2i)$

$\rm \:  =  \:  \: - 2 + 2i + 2 + 2i$

$\rm \:  =  \:  \: 4i$

Hence,

$\: \: \: \: \: \: \: \: \: \underbrace{\boxed{ \bf{ \: {(1 + i)}^{3} \: - \: {(1 - i)}^{3} \: = \: 4i \: }}}$

Additional Information :-

$\boxed{ \bf{ \: i \: = \: \sqrt{ - 1}}}$

$\boxed{ \bf{ \: \frac{1}{i} = - i}}$

$\boxed{ \bf{ \: {i}^{4} = 1}}$

$\boxed{ \bf{ \: {i}^{n} + {i}^{n + 1} + {i}^{n + 2} + {i}^{n + 3} = 0 \: \: where \: n \: is \: natural \: number}}$