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Question Number 7609 by Tawakalitu. last updated on 06/Sep/16

Divide   a^(5/2)  − 5a^2 b^(1/3)  + 10a^(3/2) 6^(2/3)  − 10ab + 5a^(1/2) b^(4/3)  − b^(5/3)   by  a^(1/2)  − b^(1/3)

$${Divide}\: \\ $$$${a}^{\mathrm{5}/\mathrm{2}} \:−\:\mathrm{5}{a}^{\mathrm{2}} {b}^{\mathrm{1}/\mathrm{3}} \:+\:\mathrm{10}{a}^{\mathrm{3}/\mathrm{2}} \mathrm{6}^{\mathrm{2}/\mathrm{3}} \:−\:\mathrm{10}{ab}\:+\:\mathrm{5}{a}^{\mathrm{1}/\mathrm{2}} {b}^{\mathrm{4}/\mathrm{3}} \:−\:{b}^{\mathrm{5}/\mathrm{3}} \\ $$$${by}\:\:{a}^{\mathrm{1}/\mathrm{2}} \:−\:{b}^{\mathrm{1}/\mathrm{3}} \\ $$

Answered by sandy_suhendra last updated on 06/Sep/16

let   a^(1/2) =x   and   b^(1/3) =y  (x^5 −5x^4 y+10x^3 y^2 −10x^2 y^3 +5xy^4 −y^5 )÷(x−y)  =(x−y)^5 ÷(x−y)  =(x−y)^4   =x^4 −4x^3 y+6x^2 y^2 −4xy^3 +y^4   =a^2 −4a^(3/2) b^(1/3) +6ab^(2/3) −4a^(1/2) b+b^(4/3)

$${let}\:\:\:{a}^{\mathrm{1}/\mathrm{2}} ={x}\:\:\:{and}\:\:\:{b}^{\mathrm{1}/\mathrm{3}} ={y} \\ $$$$\left({x}^{\mathrm{5}} −\mathrm{5}{x}^{\mathrm{4}} {y}+\mathrm{10}{x}^{\mathrm{3}} {y}^{\mathrm{2}} −\mathrm{10}{x}^{\mathrm{2}} {y}^{\mathrm{3}} +\mathrm{5}{xy}^{\mathrm{4}} −{y}^{\mathrm{5}} \right)\boldsymbol{\div}\left({x}−{y}\right) \\ $$$$=\left({x}−{y}\right)^{\mathrm{5}} \boldsymbol{\div}\left({x}−{y}\right) \\ $$$$=\left({x}−{y}\right)^{\mathrm{4}} \\ $$$$={x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{3}} {y}+\mathrm{6}{x}^{\mathrm{2}} {y}^{\mathrm{2}} −\mathrm{4}{xy}^{\mathrm{3}} +{y}^{\mathrm{4}} \\ $$$$={a}^{\mathrm{2}} −\mathrm{4}{a}^{\mathrm{3}/\mathrm{2}} {b}^{\mathrm{1}/\mathrm{3}} +\mathrm{6}{ab}^{\mathrm{2}/\mathrm{3}} −\mathrm{4}{a}^{\mathrm{1}/\mathrm{2}} {b}+{b}^{\mathrm{4}/\mathrm{3}} \\ $$

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