If (x +iy)3 =u+iv,then show that u/x+v/y=4(x2-y2). here 2 and 3 means square cube respectively.
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(x+iy)³=u+iv
or, x³+3x²iy+3xi²y²+i³y³=u+iv
or, x³+3x²iy-3xy²-iy³=u+iv
or, (x³-3xy²)+i(3x²y-y³)=u+iv
∴, x³-3xy²=u and 3x²y-y³=v
∴, u/x+v/y
=x³-3xy²/x+3x²y-y³/y
=x²-3y²+3x²-y²
=4x²-4y²
=4(x²-y²)(Proved)
or, x³+3x²iy+3xi²y²+i³y³=u+iv
or, x³+3x²iy-3xy²-iy³=u+iv
or, (x³-3xy²)+i(3x²y-y³)=u+iv
∴, x³-3xy²=u and 3x²y-y³=v
∴, u/x+v/y
=x³-3xy²/x+3x²y-y³/y
=x²-3y²+3x²-y²
=4x²-4y²
=4(x²-y²)(Proved)
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