Question

Show that $(-1+\sqrt{3}i)^{3}$ is a real number.

Answer 1

We know that

$( {x - y)}^{3} = {x}^{3} - {y}^{3} - 3 {x}^{2} y + 3x {y}^{2}$
So apply this identity

$(-1+\sqrt{3}i)^{3} = (\sqrt{3}i - 1)^{3} \\ \\ = {( \sqrt{3}i })^{3} - {(1)}^{3} - 3 {( \sqrt{3} i)}^{2} (1) + 3( \sqrt{3} i) {1}^{2} \\ \\ = 3 \sqrt{3} {i}^{3} - 1 - 9 {i}^{2} + 3 \sqrt{3} i \\ \\ we \: know \: that \\ \\ {i}^{3} = - i \\ \\ and \: \: \: {i}^{2} = - 1 \\ \\ so \\ \\ = - 3 \sqrt{3}i - 1 + 9 + 3 \sqrt{3} i \\ \\ = 8$
here at last no imaginary term left in the expression,so given expression is a real number.

Hope it helps you.

Answer 2

Answer:

8

Step-by-step explanation:

Hi,

Consider ( -1 + i√3)³

Expanding using binomial expansion, we get

= ³C₀(-1)³ + ³C₁(-1)²(i√3) + ³C₂(-1)(i√3)² + ³C₃(i√3)³

On simplifying, we get

= -1 +3i√3 - 3*3i² + (√3)³i³

But we know that i² = -1

i³ = i²*i = -1*i = -i

So , ( -1 + i√3)³ = -1 + 3√3 i -9(-1) + 3√3*(-i)

= -1 + 3√3 i + 9 - 3√3 i

= 9 - 1

= 8 which is a real number

Hence, ( -1 + i√3)³ is a real number.

Hope, it helps !