Question

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Answer 1

Question :–

▪︎ Gives that ${ \bold { \: \dfrac{ {a}^{3} + 3a {b}^{2} }{ {b}^{3} + 3 {a}^{2}b } = \dfrac{63}{62} \: }}$ , Using componendo and dividendo, find a : b

ANSWER :–

a : b = 6 : 4

EXPLANATION :–

GIVEN :–

▪︎ $\implies \: \dfrac{ {a}^{3} + 3a {b}^{2} }{ {b}^{3} + 3 {a}^{2}b } = \dfrac{63}{62}$

TO FIND :–

▪︎ a : b = ?

SOLUTION :–

☞ Componendo and Dividendo (C & D) :–

If a fraction is (a/b) = (c/d) , Now Applying (C & D) rule –

$\\ \implies \: \dfrac{a + b}{a - b } = \dfrac{c + d}{c - d}$

▪︎ Now Apply (C & D) rule in given equation –

$\\ \implies \: \dfrac{ {a}^{3} + 3a {b}^{2} + {b}^{3} + 3 {a}^{2}b }{ {a}^{3} + 3a {b}^{2} - ( {b}^{3} + 3 {a}^{2}b)} = \dfrac{63 + 62}{1}$

$\\ \implies \: \dfrac{ {a}^{3} + 3a {b}^{2} + {b}^{3} + 3 {a}^{2}b }{ {a}^{3} + 3a {b}^{2} - {b}^{3} - 3 {a}^{2}b} = \dfrac{125}{1}$

$\\ \implies \: \dfrac{ {(a + b)}^{3} }{ {(a - b)}^{3} } = \dfrac{ {(5)}^{3} }{1}$

▪︎ We should write this as –

$\\ \implies \: \dfrac{ {a + b}}{ {a - b}} = \dfrac{ {5 }}{1}$

▪︎ Now Apply (C & D) rule –

$\\ \implies \: \dfrac{ {(a + b) + (a - b)}}{ (a + b) - (a - b)} = \dfrac{ {5 + 1}}{5 - 1}$

$\\ \implies \: \dfrac{ {2a}}{2b} = \dfrac{ {6}}{4}$

$\\ \implies \: { \boxed{ \bold{ \dfrac{ {a}}{b} = \dfrac{ {6}}{4} }}}$

Hence, a : b = 6 : 4

USED FORMULA :–

(1) (a + b)³ = a³ + b³ + 3a²b + 3ab²

(2) (a - b)³ = a³ - b³ - 3a²b + 3ab²