Verified Answer :
Use formula
\(e^{ix}=\cos{x}+\sin{x}\)
\(\sqrt[3]{27(\cos{330^o}+i\sin{330^o})}
\\\sqrt[3]{27e^{i330^o} }
\\ \sqrt[3]{27} \sqrt[3]{e^{i330^o}}
\\ \sqrt[3]{3^3} \sqrt[3]{e^{i110^o\times 3}}
\\ \sqrt[3]{3^3} \sqrt[3]{(e^{i110^o})^ 3}}
\\ 3e^{i110^o}
\\3(\cos{110^o}+i\sin{110^o})\)
Since \(330^o=360^o+330^o=690^o\) and \(330^o=2\times 360^o+330^o=1050^o\) there are two more solutions:
1. \(3(\cos{230^o}+i\sin{230^o})\)
2. \(3(\cos{350^o}+i\sin{350^o})\)
Answer:
The cube roots are
\(3(cos{110^{\circ}}+isin{110^{\circ}})\)
\(3(cos{230^{\circ}}+isin{230^{\circ}})\)
\(3(cos{350^{\circ}}+isin{350^{\circ}})\)
Step-by-step explanation:
Given the expression 27(cos 330° + i sin 330°)
we have to find the cube roots of the above expression.
By Euler's formula,
\(e^{i\theta}=cos{\theta}+isin{\theta\)
We can write
\(cos{330^{\circ}}+isin{330^{\circ}}=e^{330i}\)
\(\sqrt[3]{27(cos{330^{\circ}}+isin{330^{\circ}})}\\\\=\sqrt[3]{3^3(cos{330^{\circ}}+isin{330^{\circ}})}\\\\=\sqrt[3]{3^3(e^{330i})}\\\\=\sqrt[3]{3^3(e^{110i\times3})}\\\\=\sqrt[3]{(3e^{110i})^3}\\\\=3e^{110i}=3(cos{110^{\circ}}+isin{110^{\circ}})\)
Since, 330°=360°+330°=690° and 330°=2(360°)+330°=1050°
Other two solutions are
\(3(cos{230^{\circ}}+isin{230^{\circ}})\) and \(3(cos{350^{\circ}}+isin{350^{\circ}})\)