Question

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

Answer 1

Answer:

(-1+√3i)³

by the formula (a+b)³ = a³ + 3a²b + 3ab² + b³

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

-1 + 3√3i + 3√3 - 3√3i³

-1 +3√3

which is real.

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Answer 2

Answer:

8 is a real number.

Step-by-step explanation:

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

To show : The given expression is a real number ?

Solution :

Using identity,

$(a+b)^3=a^3+b^3+3a^2b+3ab^2$

Here, a=-1 and $b=\sqrt{3i}$

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

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

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

$(-1+\sqrt{3i})^3=8$

8 is a real number.