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Question-48744

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Question Number 48744 by peter frank last updated on 28/Nov/18
Answered by Abdulhafeez Abu qatada last updated on 28/Nov/18
((x + iy))^(1/3) = a + ib x + iy = (a + ib)^3 x + iy = a^3 + 3ia^2 b − 3ab^2 −ib^3 x + iy = a^3 − 3ab^2 + i(3a^2 b − b^3 ) Compare the real and imaginary parts a^3 − 3ab^2 = x .....(1) 3a^2 b − b^3 = y .....(2) add (1) and (2) a^3 − b^3 − 3ab^2 + 3a^2 b = x + y a^3 − b^3 + 3ab(a − b) = x + y (a − b)(a^2 + b^2 + ab) + 3ab(a − b) = x + y (a − b)(a^2 + b^2 + ab + 3ab) = x + y (a − b)(a^2 + b^2 + 4ab) = x + y ....∗∗∗ Multiply (1) by a and (2) by b a^4 − 3a^2 b^2 = ax .....(3) 3a^2 b^2 − b^4 = by .....(4) add (3) and (4) a^4 − b^4 = ax + by a^2 + b^2 = ((ax + by)/(a^2 − b^2 )) Insert in ∗∗∗ (a − b)(a^2 + b^2 + 4ab) = x + y (a − b)(((ax + by)/(a^2 − b^2 )) + 4ab) = x + y (((a − b))/((a^2 − b^2 )))(ax + by + 4ab(a^2 − b^2 ) = x + y (1/(a + b))(ax + by + 4ab(a^2 − b^2 ) = x + y ax + by + 4ab(a^2 − b^2 ) = (x + y)(a + b) ax + by + 4ab(a^2 − b^2 ) = ax + by + bx + ay 4ab(a^2 − b^2 ) = bx + ay divide through by ab 4(a^2 − b^2 ) = (x/a) + (y/b) Hafeez Ayinde(Abu qatada)
$$ \\ $$$$ \\ $$$$\sqrt[{\mathrm{3}}]{{x}\:+\:{iy}}\:=\:{a}\:+\:{ib} \\ $$$${x}\:+\:{iy}\:=\:\left({a}\:+\:{ib}\right)^{\mathrm{3}} \\ $$$${x}\:+\:{iy}\:=\:{a}^{\mathrm{3}} \:+\:\mathrm{3}{ia}^{\mathrm{2}} {b}\:−\:\mathrm{3}{ab}^{\mathrm{2}} −{ib}^{\mathrm{3}} \\ $$$${x}\:+\:{iy}\:=\:{a}^{\mathrm{3}} \:−\:\mathrm{3}{ab}^{\mathrm{2}} +\:{i}\left(\mathrm{3}{a}^{\mathrm{2}} {b}\:−\:{b}^{\mathrm{3}} \right) \\ $$$${Compare}\:{the}\:{real}\:{and}\:{imaginary}\:{parts} \\ $$$${a}^{\mathrm{3}} \:−\:\mathrm{3}{ab}^{\mathrm{2}} \:=\:{x}\:…..\left(\mathrm{1}\right) \\ $$$$\mathrm{3}{a}^{\mathrm{2}} {b}\:−\:{b}^{\mathrm{3}} \:=\:{y}\:\:…..\left(\mathrm{2}\right) \\ $$$$ \\ $$$${add}\:\left(\mathrm{1}\right)\:{and}\:\left(\mathrm{2}\right) \\ $$$${a}^{\mathrm{3}} \:−\:{b}^{\mathrm{3}} \:−\:\mathrm{3}{ab}^{\mathrm{2}} \:+\:\mathrm{3}{a}^{\mathrm{2}} {b}\:=\:{x}\:+\:{y} \\ $$$${a}^{\mathrm{3}} \:−\:{b}^{\mathrm{3}} \:+\:\mathrm{3}{ab}\left({a}\:−\:{b}\right)\:=\:{x}\:+\:{y} \\ $$$$\left({a}\:−\:{b}\right)\left({a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:+\:{ab}\right)\:+\:\mathrm{3}{ab}\left({a}\:−\:{b}\right)\:=\:{x}\:+\:{y} \\ $$$$\left({a}\:−\:{b}\right)\left({a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:+\:{ab}\:+\:\mathrm{3}{ab}\right)\:=\:{x}\:+\:{y} \\ $$$$\left({a}\:−\:{b}\right)\left({a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:+\:\mathrm{4}{ab}\right)\:=\:{x}\:+\:{y}\:….\ast\ast\ast \\ $$$$ \\ $$$${Multiply}\:\left(\mathrm{1}\right)\:{by}\:{a}\:{and}\:\left(\mathrm{2}\right)\:{by}\:{b} \\ $$$$ \\ $$$${a}^{\mathrm{4}} \:−\:\mathrm{3}{a}^{\mathrm{2}} {b}^{\mathrm{2}} \:=\:{ax}\:…..\left(\mathrm{3}\right) \\ $$$$\mathrm{3}{a}^{\mathrm{2}} {b}^{\mathrm{2}} \:−\:{b}^{\mathrm{4}} \:=\:{by}\:\:…..\left(\mathrm{4}\right) \\ $$$$ \\ $$$${add}\:\left(\mathrm{3}\right)\:{and}\:\left(\mathrm{4}\right) \\ $$$${a}^{\mathrm{4}} \:−\:{b}^{\mathrm{4}} \:=\:{ax}\:+\:{by} \\ $$$${a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:=\:\frac{{ax}\:+\:{by}}{{a}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} } \\ $$$${Insert}\:{in}\:\ast\ast\ast \\ $$$$\left({a}\:−\:{b}\right)\left({a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:+\:\mathrm{4}{ab}\right)\:=\:{x}\:+\:{y} \\ $$$$\left({a}\:−\:{b}\right)\left(\frac{{ax}\:+\:{by}}{{a}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} }\:+\:\mathrm{4}{ab}\right)\:=\:{x}\:+\:{y}\: \\ $$$$\frac{\left({a}\:−\:{b}\right)}{\left({a}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} \right)}\left({ax}\:+\:{by}\:+\:\mathrm{4}{ab}\left({a}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} \right)\:=\:{x}\:+\:{y}\right. \\ $$$$\frac{\mathrm{1}}{{a}\:+\:{b}}\left({ax}\:+\:{by}\:+\:\mathrm{4}{ab}\left({a}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} \right)\:=\:{x}\:+\:{y}\right. \\ $$$${ax}\:+\:{by}\:+\:\mathrm{4}{ab}\left({a}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} \right)\:=\:\left({x}\:+\:{y}\right)\left({a}\:+\:{b}\right) \\ $$$${ax}\:+\:{by}\:+\:\mathrm{4}{ab}\left({a}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} \right)\:=\:\:{ax}\:+\:{by}\:+\:{bx}\:+\:{ay} \\ $$$$\mathrm{4}{ab}\left({a}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} \right)\:=\:{bx}\:+\:{ay} \\ $$$${divide}\:{through}\:{by}\:{ab} \\ $$$$\mathrm{4}\left({a}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} \right)\:=\:\frac{{x}}{{a}}\:+\:\frac{{y}}{{b}} \\ $$$$ \\ $$$$\boldsymbol{\mathrm{Hafeez}}\:\boldsymbol{\mathrm{Ayinde}}\left(\boldsymbol{\mathrm{Abu}}\:\boldsymbol{\mathrm{qatada}}\right) \\ $$
Commented by peter frank last updated on 28/Nov/18
thank you
$$\mathrm{thank}\:\mathrm{you} \\ $$
Answered by Kunal12588 last updated on 28/Nov/18
x+iy=a^3 +b^3 i^3 +3a^2 bi+3ab^2 i^2 ⇒x+iy=a^3 −b^3 i+3a^2 bi−3ab^2 ⇒x+iy=a^3 −3ab^2 +(3a^2 b−b^3 )i x=a^3 −3ab^2 , y=3a^2 b−b^3 ⇒(x/a)=a^2 −3b^2 , ⇒(y/b)=3a^2 −b^2 ⇒(x/a)+(y/b)=a^2 −3b^2 +3a^2 −b^2 ⇒4(a^2 −b^2 )=(x/a)+(y/b)
$${x}+{iy}={a}^{\mathrm{3}} +{b}^{\mathrm{3}} {i}^{\mathrm{3}} +\mathrm{3}{a}^{\mathrm{2}} {bi}+\mathrm{3}{ab}^{\mathrm{2}} {i}^{\mathrm{2}} \\ $$$$\Rightarrow{x}+{iy}={a}^{\mathrm{3}} −{b}^{\mathrm{3}} {i}+\mathrm{3}{a}^{\mathrm{2}} {bi}−\mathrm{3}{ab}^{\mathrm{2}} \\ $$$$\Rightarrow{x}+{iy}={a}^{\mathrm{3}} −\mathrm{3}{ab}^{\mathrm{2}} +\left(\mathrm{3}{a}^{\mathrm{2}} {b}−{b}^{\mathrm{3}} \right){i} \\ $$$${x}={a}^{\mathrm{3}} −\mathrm{3}{ab}^{\mathrm{2}} ,\:\:\:{y}=\mathrm{3}{a}^{\mathrm{2}} {b}−{b}^{\mathrm{3}} \\ $$$$\Rightarrow\frac{{x}}{{a}}={a}^{\mathrm{2}} −\mathrm{3}{b}^{\mathrm{2}} ,\:\:\Rightarrow\frac{{y}}{{b}}=\mathrm{3}{a}^{\mathrm{2}} −{b}^{\mathrm{2}} \\ $$$$\Rightarrow\frac{{x}}{{a}}+\frac{{y}}{{b}}={a}^{\mathrm{2}} −\mathrm{3}{b}^{\mathrm{2}} +\mathrm{3}{a}^{\mathrm{2}} −{b}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{4}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)=\frac{{x}}{{a}}+\frac{{y}}{{b}} \\ $$

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