Question

If $\sqrt[3]{3a-ib}=x-iy$, then $\sqrt[3]{3a+ib}=$

• A. $x+iy$
• B. $x-iy$
• C. $y+ix$
• D. $y-ix$

Answer 1

Answer: (A)

$x+iy$

We have $\sqrt[3]{3a-ib}=x-iy$

$\Rightarrow a-ib={(x-iy)}^3=x^3-3x^2.iy+3x{\left(iy\right)}^2-{\left(iy\right)}^3$

$=(x^3-3x{y}^2)-i(3x^{2}y-y^3)$

$\therefore a+ib=(x^3-3x{y}^2)+i(3x^{2}y-y^3)$

$=x^3+3x^2.\left(iy\right)+3x{\left(iy\right)}^2+{\left(iy\right)}^3={(x+iy)}^3$

$\therefore \sqrt[3]{3a+ib}=x+iy$