Question

Show that (-1+√3i)^3 is a real number.

Answer 1

(√3 i - 1)³

using the formula

(a - b)³ = a³ - b³ - 3 a²b + 3 ab²

expanding the cube

(√3)³ i³ - 1³ - 3 (√3 i)² (1) + 3 (√3 i) (1)²

- 3√3 i - 1 - 9 i² + 3√3 i

- 3√3 i - 1 - 9 (-1) + 3√3 i                     (Since i² = - 1)

- 3√3 i - 1 + 9 + 3√3 i

8

hence proved

Answer 2

Answer:

Yes (-1+√3i)^3 is a real number that is 8

Explanation:

We know the (a + b)³ = (a + b)(a + b)(a + b)

In (-1+√3i)³ , a = -1 & b = i√3

(-1+√3i)³ = (-1 + i√3)(-1 + i√3)(-1 + i√3)

(-1+√3i)³ = (1 - 3 - 2i√3)(-1 + i√3)

(-1+√3i)³ = (-2 - 2i√3)(-1 + i√3)

(-1+√3i)³ = -2(1 + i√3)(-1 + i√3)

(-1+√3i)³ = -2 (-1 - 3)

(-1 + i√3)³ = -2 (-4)

(-1 + i√3)³ = 8

8 is real number.

That's the final answer. I hope it will help you.