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Question Number 119996 by behi83417@gmail.com last updated on 28/Oct/20

 { ((x^3 +y^2 =a)),((x^2 +y^3 =b)) :}   [solve for:x,y,a≠b∈R]

$$\begin{cases}{\boldsymbol{{x}}^{\mathrm{3}} +\boldsymbol{{y}}^{\mathrm{2}} =\boldsymbol{{a}}}\\{\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{3}} =\boldsymbol{{b}}}\end{cases}\:\:\:\left[\boldsymbol{{solve}}\:\boldsymbol{{for}}:\mathrm{x},\mathrm{y},\mathrm{a}\neq\mathrm{b}\in\boldsymbol{\mathrm{R}}\right] \\ $$

Answered by TANMAY PANACEA last updated on 28/Oct/20

x^6 =(a−y^2 )^2   x^6 =(b−y^3 )^3   a^2 −2ay^2 +y^4 =b^3 −3b^2 y^3 +3by^6 −y^9   y^9 −3by^6 +y^4 +3b^3 y^3 −2ay^2 −b^3 =0  Wait pls

$${x}^{\mathrm{6}} =\left({a}−{y}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{6}} =\left({b}−{y}^{\mathrm{3}} \right)^{\mathrm{3}} \\ $$$${a}^{\mathrm{2}} −\mathrm{2}{ay}^{\mathrm{2}} +{y}^{\mathrm{4}} ={b}^{\mathrm{3}} −\mathrm{3}{b}^{\mathrm{2}} {y}^{\mathrm{3}} +\mathrm{3}{by}^{\mathrm{6}} −{y}^{\mathrm{9}} \\ $$$${y}^{\mathrm{9}} −\mathrm{3}{by}^{\mathrm{6}} +{y}^{\mathrm{4}} +\mathrm{3}{b}^{\mathrm{3}} {y}^{\mathrm{3}} −\mathrm{2}{ay}^{\mathrm{2}} −{b}^{\mathrm{3}} =\mathrm{0} \\ $$$$\boldsymbol{{W}}{ait}\:{pls} \\ $$$$ \\ $$$$ \\ $$

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