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Question-205884

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Question Number 205884 by Abdullahrussell last updated on 01/Apr/24
Commented by Frix last updated on 01/Apr/24
a^3 +b^3
$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} \\ $$
Answered by Rasheed.Sindhi last updated on 04/Apr/24
(((a^4 +2ab^3 )^3 −(ab^4 +2a^3 b)^3 )/((a^3 −b^3 )^3 )) =((A^3 −B^3 )/((a^3 −b^3 )^3 )) =(((A−B)( (A−B)^2 +3AB ))/((a−b)^3 (a^2 +ab+b^2 )^3 )) =((A−B)/((a−b)^3 (a^2 +ab+b^2 )^3 ))∙( (A−B)^2 +3AB ) A−B=a^4 −b^4 +2ab^3 −2a^3 b =(a^2 −b^2 )(a^2 +b^2 )−2ab(a^2 −b^2 ) =(a^2 −b^2 )(a^2 +b^2 −2ab) =(a−b)^3 (a+b) ((A−B)/((a−b)^3 (a^2 +ab+b^2 )^3 ))=(((a−b)^3 (a+b))/((a−b)^3 (a^2 +ab+b^2 )^3 )) =((a+b)/((a^2 +ab+b^2 )^3 )) ( (A−B)^2 +3AB ) =(((a−b)^3 (a+b))^2 +3(a^4 +2ab^3 )(b^4 +2a^3 b)) =((a−b)^6 (a+b)^2 +3(a^4 b^4 +4a^4 b^4 +2ab^7 +2a^7 b)) =((a−b)^6 (a+b)^2 +15a^4 b^4 +6ab(a^6 +b^6 )) =(a^2 −ab+b^2 )(a^2 +ab+b^2 )^3 ((A−B)/((a−b)^3 (a^2 +ab+b^2 )^3 ))∙( (A−B)^2 +3AB ) =((a+b)/((a^2 +ab+b^2 )^3 ))∙(a^2 −ab+b^2 )(a^2 +ab+b^2 )^3 =(a+b)(a^2 −ab+b^2 )=a^3 +b^3
$$\frac{\left({a}^{\mathrm{4}} +\mathrm{2}{ab}^{\mathrm{3}} \right)^{\mathrm{3}} −\left({ab}^{\mathrm{4}} +\mathrm{2}{a}^{\mathrm{3}} {b}\right)^{\mathrm{3}} }{\left({a}^{\mathrm{3}} −{b}^{\mathrm{3}} \right)^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{A}^{\mathrm{3}} −\mathrm{B}^{\mathrm{3}} }{\left({a}^{\mathrm{3}} −{b}^{\mathrm{3}} \right)^{\mathrm{3}} } \\ $$$$=\frac{\left(\mathrm{A}−\mathrm{B}\right)\left(\:\left(\mathrm{A}−\mathrm{B}\right)^{\mathrm{2}} +\mathrm{3AB}\:\right)}{\left({a}−{b}\right)^{\mathrm{3}} \left({a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} \right)^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{A}−\mathrm{B}}{\left({a}−{b}\right)^{\mathrm{3}} \left({a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} \right)^{\mathrm{3}} }\centerdot\left(\:\left(\mathrm{A}−\mathrm{B}\right)^{\mathrm{2}} +\mathrm{3AB}\:\right) \\ $$$$\mathrm{A}−\mathrm{B}={a}^{\mathrm{4}} −{b}^{\mathrm{4}} +\mathrm{2}{ab}^{\mathrm{3}} −\mathrm{2}{a}^{\mathrm{3}} {b} \\ $$$$\:\:\:\:\:\:\:\:=\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)−\mathrm{2}{ab}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:\:=\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\left({a}−{b}\right)^{\mathrm{3}} \left({a}+{b}\right) \\ $$$$\frac{\mathrm{A}−\mathrm{B}}{\left({a}−{b}\right)^{\mathrm{3}} \left({a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} \right)^{\mathrm{3}} }=\frac{\cancel{\left({a}−{b}\right)^{\mathrm{3}} }\left({a}+{b}\right)}{\cancel{\left({a}−{b}\right)^{\mathrm{3}} }\left({a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} \right)^{\mathrm{3}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{a}+{b}}{\left({a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} \right)^{\mathrm{3}} } \\ $$$$\left(\:\left(\mathrm{A}−\mathrm{B}\right)^{\mathrm{2}} +\mathrm{3AB}\:\right) \\ $$$$=\left(\left(\left({a}−{b}\right)^{\mathrm{3}} \left({a}+{b}\right)\right)^{\mathrm{2}} +\mathrm{3}\left({a}^{\mathrm{4}} +\mathrm{2}{ab}^{\mathrm{3}} \right)\left({b}^{\mathrm{4}} +\mathrm{2}{a}^{\mathrm{3}} {b}\right)\right) \\ $$$$=\left(\left({a}−{b}\right)^{\mathrm{6}} \left({a}+{b}\right)^{\mathrm{2}} +\mathrm{3}\left({a}^{\mathrm{4}} {b}^{\mathrm{4}} +\mathrm{4}{a}^{\mathrm{4}} {b}^{\mathrm{4}} +\mathrm{2}{ab}^{\mathrm{7}} +\mathrm{2}{a}^{\mathrm{7}} {b}\right)\right) \\ $$$$=\left(\left({a}−{b}\right)^{\mathrm{6}} \left({a}+{b}\right)^{\mathrm{2}} +\mathrm{15}{a}^{\mathrm{4}} {b}^{\mathrm{4}} +\mathrm{6}{ab}\left({a}^{\mathrm{6}} +{b}^{\mathrm{6}} \right)\right) \\ $$$$=\left({a}^{\mathrm{2}} −{ab}+{b}^{\mathrm{2}} \right)\left({a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} \right)^{\mathrm{3}} \\ $$$$\frac{\mathrm{A}−\mathrm{B}}{\left({a}−{b}\right)^{\mathrm{3}} \left({a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} \right)^{\mathrm{3}} }\centerdot\left(\:\left(\mathrm{A}−\mathrm{B}\right)^{\mathrm{2}} +\mathrm{3AB}\:\right) \\ $$$$=\frac{{a}+{b}}{\cancel{\left({a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} \right)^{\mathrm{3}} }}\centerdot\left({a}^{\mathrm{2}} −{ab}+{b}^{\mathrm{2}} \right)\cancel{\left({a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} \right)^{\mathrm{3}} } \\ $$$$=\left({a}+{b}\right)\left({a}^{\mathrm{2}} −{ab}+{b}^{\mathrm{2}} \right)={a}^{\mathrm{3}} +{b}^{\mathrm{3}} \\ $$
Answered by A5T last updated on 03/Apr/24
=((a^(12) +8a^3 b^9 +6a^9 b^3 −b^(12) −8a^9 b^3 −6a^3 b^9 )/((a^3 −b^3 )^3 )) =((a^(12) −b^(12) +2a^3 b^9 −2a^9 b^3 )/((a^3 −b^3 )^3 ))=(((a^6 −b^6 )(a^6 +b^6 )−2a^3 b^3 (−b^6 +a^6 ))/((a^3 −b^3 )^3 )) =(((a^3 +b^3 )(a^3 −b^3 )(a^6 +b^6 −2a^3 b^3 ))/((a^3 −b^3 )^3 ))=a^3 +b^3
$$=\frac{{a}^{\mathrm{12}} +\mathrm{8}{a}^{\mathrm{3}} {b}^{\mathrm{9}} +\mathrm{6}{a}^{\mathrm{9}} {b}^{\mathrm{3}} −{b}^{\mathrm{12}} −\mathrm{8}{a}^{\mathrm{9}} {b}^{\mathrm{3}} −\mathrm{6}{a}^{\mathrm{3}} {b}^{\mathrm{9}} }{\left({a}^{\mathrm{3}} −{b}^{\mathrm{3}} \right)^{\mathrm{3}} } \\ $$$$=\frac{{a}^{\mathrm{12}} −{b}^{\mathrm{12}} +\mathrm{2}{a}^{\mathrm{3}} {b}^{\mathrm{9}} −\mathrm{2}{a}^{\mathrm{9}} {b}^{\mathrm{3}} }{\left({a}^{\mathrm{3}} −{b}^{\mathrm{3}} \right)^{\mathrm{3}} }=\frac{\left({a}^{\mathrm{6}} −{b}^{\mathrm{6}} \right)\left({a}^{\mathrm{6}} +{b}^{\mathrm{6}} \right)−\mathrm{2}{a}^{\mathrm{3}} {b}^{\mathrm{3}} \left(−{b}^{\mathrm{6}} +{a}^{\mathrm{6}} \right)}{\left({a}^{\mathrm{3}} −{b}^{\mathrm{3}} \right)^{\mathrm{3}} } \\ $$$$=\frac{\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} \right)\left({a}^{\mathrm{3}} −{b}^{\mathrm{3}} \right)\left({a}^{\mathrm{6}} +{b}^{\mathrm{6}} −\mathrm{2}{a}^{\mathrm{3}} {b}^{\mathrm{3}} \right)}{\left({a}^{\mathrm{3}} −{b}^{\mathrm{3}} \right)^{\mathrm{3}} }={a}^{\mathrm{3}} +{b}^{\mathrm{3}} \\ $$

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