[tex] \large\underline{\sf{Solution-}}[/tex]
Given that,
[tex]\rm \longmapsto\: {(x + iy)}^{3} = y + iv[/tex]
[tex]\rm \longmapsto\: {x}^{3} + 3 {x}^{2}(iy) + 3x {(iy)}^{2} + {(iy)}^{3} = y + iv[/tex]
[tex]\rm \longmapsto\: {x}^{3} + 3 {x}^{2}yi + 3x {i}^{2} {y}^{2} + {i}^{3} {y}^{3} = y + iv[/tex]
We know,
[tex] \purple{\rm \longmapsto\:\boxed{\tt{ {i}^{2} = - 1 \: \: \: and \: \: \: {i}^{3} = - i }}} \\ [/tex]
So, using this, we get
[tex]\rm :\longmapsto\: {x}^{3} + 3 {x}^{2}yi - 3x{y}^{2} - i{y}^{3} = y + iv[/tex]
[tex]\rm \longmapsto\: ({x}^{3} - 3x {y}^{2}) + i(3{x}^{2}y- {y}^{3})= y + iv[/tex]
On comparing real and Imaginary parts, we get
[tex]\rm \longmapsto\:y = {x}^{3} - 3 {xy}^{2} [/tex]
and
[tex]\rm \longmapsto\:v = {3x}^{2}y - {y}^{3} [/tex]
Now, Consider
[tex]\rm \longmapsto\:\dfrac{y}{x} + \dfrac{v}{y} [/tex]
[tex]\rm \: = \: \dfrac{ {x}^{3} - {3xy}^{2}}{x} + \dfrac{ {3yx}^{2} - {y}^{3} }{y} [/tex]
[tex]\rm \: = \: \dfrac{ x({x}^{2} - {3y}^{2})}{x} + \dfrac{y({3x}^{2} - {y}^{2})}{y} [/tex]
[tex]\rm \: = \: {x}^{2} - {3y}^{2} + {3x}^{2} - {y}^{2} [/tex]
[tex]\rm \: = \: 4{x}^{2} - {4y}^{2}[/tex]
[tex]\rm \: = \: 4({x}^{2} - {y}^{2})[/tex]
Hence, Proved
