Question

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

Answer 1

Hint: For the above question we will have to know about the real number. Real numbers are simply the combination of rational numbers and irrational numbers. They can be positive, negative or zero. All the natural numbers, decimal numbers and fractions come under this category.

Complete Step-by-Step solution:
\[{{(a+b)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab(a+b)\]
Now, using the above formula we get:
\[\begin{align}
  & {{\left( -1+\sqrt{3}i \right)}^{3}}=-1-3\sqrt{3}i-3\sqrt{3}i\left( -1+\sqrt{3}i \right) \\
 & \text{ =}-1-3\sqrt{3}i+3\sqrt{3}i+9 \\
 & \text{ = 9}-1 \\
 & \text{ = 8} \\
\end{align}\]
On expanding the ${{\left( -1+\sqrt{3}i \right)}^{3}}$we get 8, which is a real number.
Hence, it is proved that the given expression in the above question is a real number.

Note: Just be careful while doing calculation as there is a chance that you might make a mistake while expanding the ${{\left( -1+\sqrt{3}i \right)}^{3}}$and you will get the incorrect solution.
Also remember the formula of \[{{(a+b)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab(a+b)\].
Also there is a fact about the zero that is considered as both real and an imaginary number. As we know, imaginary numbers are the square root of non-positive real numbers. And since 0 is also a non-negative number, therefore it fulfils the criteria of the imaginary number, whereas 0 is also a rational number, which is defined in a number line and hence a real number.