Question

plz answer it guys l need help​

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

1.

A complex number is said to be real number if

Imaginary part of the number = 0

2.

${(a + b)}^{3} = {a}^{3} + 3 {a}^{2}b + 3a {b}^{2} + {b}^{3}$

3.

${i}^{2} = - 1$

TO PROVE

$\sf{ {( - 1 + \sqrt{3} i)}^{3} \: is \: a \: real \: number \: }$

PROOF

${( - 1 + \sqrt{3} i)}^{3}$

$= {( - 1)}^{3} + 3 {( - 1)}^{2} \sqrt{3}i + 3( - 1) {( \sqrt{3}i) }^{2} + {( \sqrt{3}i) }^{3}$

$= - 1 + 3\sqrt{3}i - 3 \times 3 \times {i}^{2} + 3 \sqrt{3} {i}^{3}$

$= - 1 + 3\sqrt{3}i + 9 - 3 \sqrt{3} {i}$

$= 8$

Which is purely real number

Answer 2

Answer:

(−1+

3

i)

3

= {( - 1)}^{3} + 3 {( - 1)}^{2} \sqrt{3}i + 3( - 1) {( \sqrt{3}i) }^{2} + {( \sqrt{3}i) }^{3}=(−1)

3

+3(−1)

2

3

i+3(−1)(

3

i)

2

+(

3

i)

3

= - 1 + 3\sqrt{3}i - 3 \times 3 \times {i}^{2} + 3 \sqrt{3} {i}^{3}=−1+3

3

i−3×3×i

2

+3

3

i

3

= - 1 + 3\sqrt{3}i + 9 - 3 \sqrt{3} {i}=−1+3

3

i+9−3

3

i

= 8=8

Which is purely real number