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Question Number 159394 by azadsir last updated on 16/Nov/21

if a^3 −b^3 =513, ab=54 than, a−b = ?

$$\mathrm{if}\:\mathrm{a}^{\mathrm{3}} −\mathrm{b}^{\mathrm{3}} =\mathrm{513},\:\mathrm{ab}=\mathrm{54} \\ $$$$\:\mathrm{than},\:\mathrm{a}−\mathrm{b}\:=\:? \\ $$

Commented by cortano last updated on 16/Nov/21

a^3 −b^3 =(a−b)(a^2 +b^2 +ab)=513 (a−b)((a−b)^2 +3ab))=513 (a−b)((a−b)^2 +162)=513 x^3 +162x−513=0 (x−3)(x^2 +3x+171)=0 ⇒x=3∈R ; a−b=3

$${a}^{\mathrm{3}} −{b}^{\mathrm{3}} =\left({a}−{b}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{ab}\right)=\mathrm{513} \\ $$$$\left.\left({a}−{b}\right)\left(\left({a}−{b}\right)^{\mathrm{2}} +\mathrm{3}{ab}\right)\right)=\mathrm{513} \\ $$$$\left({a}−{b}\right)\left(\left({a}−{b}\right)^{\mathrm{2}} +\mathrm{162}\right)=\mathrm{513} \\ $$$${x}^{\mathrm{3}} +\mathrm{162}{x}−\mathrm{513}=\mathrm{0} \\ $$$$\left({x}−\mathrm{3}\right)\left({x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{171}\right)=\mathrm{0} \\ $$$$\:\Rightarrow{x}=\mathrm{3}\in\mathbb{R}\:;\:{a}−{b}=\mathrm{3} \\ $$$$ \\ $$

Commented by azadsir last updated on 16/Nov/21

Thank you

$$\mathrm{Thank}\:\mathrm{you} \\ $$

Answered by Rasheed.Sindhi last updated on 16/Nov/21

Another Way: b=((54)/a) a^3 −(((54)/a))^3 =513 a^6 −513a^3 −54^3 =0 a^3 =((513±(√(513^2 +4.54^3 )))/2)=729,−216 a=9,−6⇒b=((54)/9),((54)/(−6))=6,−9 a−b=9−6, −6−(−9)=3,3 ∴ a−b=3

$$\underline{\mathbb{A}\mathrm{nother}\:\mathbb{W}\mathrm{ay}:} \\ $$$$\mathrm{b}=\frac{\mathrm{54}}{\mathrm{a}} \\ $$$$\mathrm{a}^{\mathrm{3}} −\left(\frac{\mathrm{54}}{\mathrm{a}}\right)^{\mathrm{3}} =\mathrm{513} \\ $$$$\mathrm{a}^{\mathrm{6}} −\mathrm{513a}^{\mathrm{3}} −\mathrm{54}^{\mathrm{3}} =\mathrm{0} \\ $$$$\mathrm{a}^{\mathrm{3}} =\frac{\mathrm{513}\pm\sqrt{\mathrm{513}^{\mathrm{2}} +\mathrm{4}.\mathrm{54}^{\mathrm{3}} }}{\mathrm{2}}=\mathrm{729},−\mathrm{216} \\ $$$$\mathrm{a}=\mathrm{9},−\mathrm{6}\Rightarrow\mathrm{b}=\frac{\mathrm{54}}{\mathrm{9}},\frac{\mathrm{54}}{−\mathrm{6}}=\mathrm{6},−\mathrm{9} \\ $$$$\mathrm{a}−\mathrm{b}=\mathrm{9}−\mathrm{6},\:−\mathrm{6}−\left(−\mathrm{9}\right)=\mathrm{3},\mathrm{3} \\ $$$$\therefore\:\mathrm{a}−\mathrm{b}=\mathrm{3} \\ $$

Answered by Rasheed.Sindhi last updated on 16/Nov/21

Still Another Way... ab=54⇒b=((54)/a) a−((54)/a)=c=? (a−((54)/a))^3 =c^3 a^3 −((54^3 )/a^3 )−3(a)(((54)/a))(a−((54)/a))=c^3 513−3(a)(((54)/a))(c)=c^3 c^3 +162c−513=0 (c−3)(c^2 +3x+171)=0 c−3=0 ⇒c=3 c=a−b=3

$$\mathbb{S}\mathrm{till}\:\mathbb{A}\mathrm{nother}\:\mathbb{W}\mathrm{ay}... \\ $$$${ab}=\mathrm{54}\Rightarrow{b}=\frac{\mathrm{54}}{{a}} \\ $$$${a}−\frac{\mathrm{54}}{{a}}={c}=? \\ $$$$\left({a}−\frac{\mathrm{54}}{{a}}\right)^{\mathrm{3}} ={c}^{\mathrm{3}} \\ $$$${a}^{\mathrm{3}} −\frac{\mathrm{54}^{\mathrm{3}} }{{a}^{\mathrm{3}} }−\mathrm{3}\left({a}\right)\left(\frac{\mathrm{54}}{{a}}\right)\left({a}−\frac{\mathrm{54}}{{a}}\right)={c}^{\mathrm{3}} \\ $$$$\mathrm{513}−\mathrm{3}\left({a}\right)\left(\frac{\mathrm{54}}{{a}}\right)\left({c}\right)={c}^{\mathrm{3}} \\ $$$${c}^{\mathrm{3}} +\mathrm{162}{c}−\mathrm{513}=\mathrm{0} \\ $$$$\left({c}−\mathrm{3}\right)\left({c}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{171}\right)=\mathrm{0} \\ $$$${c}−\mathrm{3}=\mathrm{0}\:\Rightarrow{c}=\mathrm{3} \\ $$$${c}={a}−{b}=\mathrm{3} \\ $$

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