Question 1060987
Q.1 {{{(x+iy)^("1 / 3")=a+ib}}} <---> {{{x+iy=(a+ib)^3}}}
{{{x+iy=(a+ib)^3}}}
{{{x+iy=a^3+3a^2*(ib)+3a*(ib)^2+(ib)^3}}}
{{{x+iy=a^3+3ba^2*i+3a*(-b^2)-b^3*i}}}
{{{x+iy=(a^3-3ab^2)+(3ba^2-b^3)*i}}}
{{{x+iy=a(a^2-3b^2)+b(3a^2-b^2)*i}}} ---> {{{system(x=a(a^2-3b^2),y=b(3a^2-b^2))}}} ---> {{{system(x/a=a^2-3b^2,y/b=3a^2-b^2))}}}
So,
{{{x/a+y/b=a^2-3b^2+3a^2-b^2}}}
{{{x/a+y/b=4a^2-4b^2}}}
{{{x/a+y/b=4(a^2-b^2)}}}


Q.2 {{{(x-iy)(3+5i)=3x+5x*i-3y*i+5y=(3x+5y)+(5x-3y)*i}}}
The conjugate of {{{-6-24i}}} is {{{-6+24i}}} .
{{{(3x+5y)+(5x-3y)*i=-6+24i}}} --> {{{system(3x+5y=-6,5x-3y=24)}}} --> {{{highlight(system(x=3,y=-3))}}}
There are many ways to solve the system. For example,
we could start by multiplying the first equation times ({{{-6}}} , the second equation times {{{3}}} and adding them to get
{{{-15x-25y+15x-9y=30+72}}} --> {{{-34y=102}}} --> {{{y=102/(-34)}}} --> {{{y=-3)}}} .
Then, substituting {{{-3}}} for {{{x}}} in {{{3x+5y=-6}}}, we get
{{{3x+5(-3)=-6}}} --> {{{3x-15=-6}}} --> {{{3x=15-6}}} --> {{{3x=9}}} --> {{{x=9/3}}} --> {{{x=3}}} .