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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

Thanks sir. I appreciate.

$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$

Commented by Spillover last updated on 05/Jul/24

perfect

$${perfect} \\ $$

Answered by MATHEMATICSAM last updated on 04/Jul/24

((a^3 + 3ab^2 )/(b^3 + 3a^2 b)) = ((63)/(62)) Using comp and div ⇒ ((a^3 + 3ab^2 + 3a^2 b + b^3 )/(a^3 + 3ab^2 − b^3 − 3a^2 b)) = ((63 + 62)/(63 − 62)) = 125 ⇒ (((a + b)^3 )/((a − b)^3 )) = 125 ⇒ ((a + b)/(a − b)) = 5 Using comp and div ⇒ (a/b) = ((5 + 1)/(5 − 1)) = (6/4) = (3/2)

$$\frac{{a}^{\mathrm{3}} \:+\:\mathrm{3}{ab}^{\mathrm{2}} }{{b}^{\mathrm{3}} \:+\:\mathrm{3}{a}^{\mathrm{2}} {b}}\:=\:\frac{\mathrm{63}}{\mathrm{62}} \\ $$$$\mathrm{Using}\:\mathrm{comp}\:\mathrm{and}\:\mathrm{div} \\ $$$$\Rightarrow\:\frac{{a}^{\mathrm{3}} \:+\:\mathrm{3}{ab}^{\mathrm{2}} \:+\:\mathrm{3}{a}^{\mathrm{2}} {b}\:+\:{b}^{\mathrm{3}} }{{a}^{\mathrm{3}} \:+\:\mathrm{3}{ab}^{\mathrm{2}} \:−\:{b}^{\mathrm{3}} \:−\:\mathrm{3}{a}^{\mathrm{2}} {b}}\:=\:\frac{\mathrm{63}\:+\:\mathrm{62}}{\mathrm{63}\:−\:\mathrm{62}}\:=\:\mathrm{125} \\ $$$$\Rightarrow\:\frac{\left({a}\:+\:{b}\right)^{\mathrm{3}} }{\left({a}\:−\:{b}\right)^{\mathrm{3}} }\:=\:\mathrm{125} \\ $$$$\Rightarrow\:\frac{{a}\:+\:{b}}{{a}\:−\:{b}}\:=\:\mathrm{5} \\ $$$$\mathrm{Using}\:\mathrm{comp}\:\mathrm{and}\:\mathrm{div} \\ $$$$\Rightarrow\:\frac{{a}}{{b}}\:=\:\frac{\mathrm{5}\:+\:\mathrm{1}}{\mathrm{5}\:−\:\mathrm{1}}\:=\:\frac{\mathrm{6}}{\mathrm{4}}\:=\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$

Commented by Tawa11 last updated on 04/Jul/24

Thanks sir. I appreciate.

$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$

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