Q.If $(x - iy)^{1/3} = a - ib$, then show that: $\dfrac{x}{a} + \dfrac{y}{b} = 4(a^2 - b^2)$.

Cube both sides to remove the cube root, expand using the binomial theorem for complex numbers, equate real and imaginary parts, then combine $x/a$ and $y/b$.

Given $(x-iy)^{1/3}=a-ib$. Cubing both sides:

$x-iy = (a-ib)^3$

Expand using $(A-B)^3=A^3-3A^2B+3AB^2-B^3$ with $A=a,B=ib$:

$(a-ib)^3 = a^3 - 3a^2(ib) + 3a(ib)^2 - (ib)^3$

$= a^3 - 3a^2bi + 3ab^2i^2 - i^3b^3$

Since $i^2=-1$ and $i^3=-i$:

$= a^3 - 3a^2bi - 3ab^2 + ib^3 = (a^3-3ab^2) + i(b^3-3a^2b)$

So:

$x - iy = (a^3-3ab^2) + i(b^3-3a^2b)$

Equating real and imaginary parts: …