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a-b-c-d-a-3-b-3-625-c-3-d-3-1-Find-a-b-c-d-




Question Number 207394 by hardmath last updated on 13/May/24
(a/b)  =  (c/d)  a^3  − b^3  = 625  c^3  − d^3  = 1  Find:   a,b,c,d = ?
$$\frac{\mathrm{a}}{\mathrm{b}}\:\:=\:\:\frac{\mathrm{c}}{\mathrm{d}} \\ $$$$\mathrm{a}^{\mathrm{3}} \:−\:\mathrm{b}^{\mathrm{3}} \:=\:\mathrm{625} \\ $$$$\mathrm{c}^{\mathrm{3}} \:−\:\mathrm{d}^{\mathrm{3}} \:=\:\mathrm{1} \\ $$$$\mathrm{Find}:\:\:\:\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\:=\:? \\ $$
Answered by mr W last updated on 13/May/24
(a/b)=(c/d)=k, say  a=kb, c=kd with k≠1  b^3 (k^3 −1)=625 ⇒b=(((625)/(k^3 −1)))^(1/3)  ⇒a=k(((625)/(k^3 −1)))^(1/3)   d^3 (k^3 −1)=1 ⇒d=(1/( ((k^3 −1))^(1/3) )) ⇒c=(k/( ((k^3 −1))^(1/3) ))  3 equations for 4 variables, so  there is no unique solution.
$$\frac{{a}}{{b}}=\frac{{c}}{{d}}={k},\:{say} \\ $$$${a}={kb},\:{c}={kd}\:{with}\:{k}\neq\mathrm{1} \\ $$$${b}^{\mathrm{3}} \left({k}^{\mathrm{3}} −\mathrm{1}\right)=\mathrm{625}\:\Rightarrow{b}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{625}}{{k}^{\mathrm{3}} −\mathrm{1}}}\:\Rightarrow{a}={k}\sqrt[{\mathrm{3}}]{\frac{\mathrm{625}}{{k}^{\mathrm{3}} −\mathrm{1}}} \\ $$$${d}^{\mathrm{3}} \left({k}^{\mathrm{3}} −\mathrm{1}\right)=\mathrm{1}\:\Rightarrow{d}=\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{k}^{\mathrm{3}} −\mathrm{1}}}\:\Rightarrow{c}=\frac{{k}}{\:\sqrt[{\mathrm{3}}]{{k}^{\mathrm{3}} −\mathrm{1}}} \\ $$$$\mathrm{3}\:{equations}\:{for}\:\mathrm{4}\:{variables},\:{so} \\ $$$${there}\:{is}\:{no}\:{unique}\:{solution}. \\ $$

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