Question Number 209223 by Tawa11 last updated on 04/Jul/24
Answered by efronzo1 last updated on 04/Jul/24
$$\:\:\:\:\cancel{\underbrace{\Subset}} \\ $$
Commented by Tawa11 last updated on 04/Jul/24
$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$
Commented by Spillover last updated on 05/Jul/24
$${perfect} \\ $$
Answered by MATHEMATICSAM last updated on 04/Jul/24
$$\frac{{a}^{\mathrm{3}} \:+\:\mathrm{3}{ab}^{\mathrm{2}} }{{b}^{\mathrm{3}} \:+\:\mathrm{3}{a}^{\mathrm{2}} {b}}\:=\:\frac{\mathrm{63}}{\mathrm{62}} \\ $$$$\mathrm{Using}\:\mathrm{comp}\:\mathrm{and}\:\mathrm{div} \\ $$$$\Rightarrow\:\frac{{a}^{\mathrm{3}} \:+\:\mathrm{3}{ab}^{\mathrm{2}} \:+\:\mathrm{3}{a}^{\mathrm{2}} {b}\:+\:{b}^{\mathrm{3}} }{{a}^{\mathrm{3}} \:+\:\mathrm{3}{ab}^{\mathrm{2}} \:−\:{b}^{\mathrm{3}} \:−\:\mathrm{3}{a}^{\mathrm{2}} {b}}\:=\:\frac{\mathrm{63}\:+\:\mathrm{62}}{\mathrm{63}\:−\:\mathrm{62}}\:=\:\mathrm{125} \\ $$$$\Rightarrow\:\frac{\left({a}\:+\:{b}\right)^{\mathrm{3}} }{\left({a}\:−\:{b}\right)^{\mathrm{3}} }\:=\:\mathrm{125} \\ $$$$\Rightarrow\:\frac{{a}\:+\:{b}}{{a}\:−\:{b}}\:=\:\mathrm{5} \\ $$$$\mathrm{Using}\:\mathrm{comp}\:\mathrm{and}\:\mathrm{div} \\ $$$$\Rightarrow\:\frac{{a}}{{b}}\:=\:\frac{\mathrm{5}\:+\:\mathrm{1}}{\mathrm{5}\:−\:\mathrm{1}}\:=\:\frac{\mathrm{6}}{\mathrm{4}}\:=\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$
Commented by Tawa11 last updated on 04/Jul/24
$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$