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Question Number 137772 by peter frank last updated on 06/Apr/21

	Answered by bemath last updated on 06/Apr/21
	$⇔ (((2^n )^3 +(3^n )^3 )/((2^n )^2 (3)^n +2^n .(3^n )^2 )) = (7/6) let { ((2^n =u)),((3^n =v)) :} ⇔ ((u^3 +v^3 )/(u^2 v+uv^2 )) = (7/6) ⇔ 6u^3 +6v^3 = 7u^2 v+7uv^2 ⇒u^2 (6u−7v)=v^2 (7u−6v) ⇒(u^2 /v^2 ) = ((7u−6v)/(6u−7u)) =((7((u/v))−6)/(6((u/v))−7)) let (u/v) =ℓ ⇒ℓ^2 = ((7ℓ−6)/(6ℓ−7)) ⇔ 6ℓ^3 −7ℓ^2 −7ℓ+6=0 (ℓ+1)(6ℓ^2 −13ℓ+6)=0 ℓ=−1 rejected (ℓ−(3/2))(ℓ−(2/3))=0 now it easy to solve n=1 or n=−1$
	$\Leftrightarrow \frac{{(2^{n})}^{3} + {(3^{n})}^{3}}{{(2^{n})}^{2} {(3)}^{n} + 2^{n} . {(3^{n})}^{2}} = \frac{7}{6}$ $l e t {\begin{cases} 2^{n} = u \\ 3^{n} = v \end{cases}$ $\Leftrightarrow \frac{u^{3} + v^{3}}{u^{2} v + {u v}^{2}} = \frac{7}{6}$ $\Leftrightarrow 6 u^{3} + 6 v^{3} = 7 u^{2} v + 7 {u v}^{2}$ $\Rightarrow u^{2} (6 u - 7 v) = v^{2} (7 u - 6 v)$ $\Rightarrow \frac{u^{2}}{v^{2}} = \frac{7 u - 6 v}{6 u - 7 u} = \frac{7 (\frac{u}{v}) - 6}{6 (\frac{u}{v}) - 7}$ $l e t \frac{u}{v} = ℓ \Rightarrow ℓ^{2} = \frac{7 ℓ - 6}{6 ℓ - 7}$ $\Leftrightarrow 6 ℓ^{3} - 7 ℓ^{2} - 7 ℓ + 6 = 0$ $(ℓ + 1) (6 ℓ^{2} - 13 ℓ + 6) = 0$ $ℓ = - 1 r e j e c t e d$ $(ℓ - \frac{3}{2}) (ℓ - \frac{2}{3}) = 0$ $n o w i t e a s y t o s o l v e$ $n = 1 o r n = - 1$
	Answered by Rasheed.Sindhi last updated on 07/Apr/21

	$\frac{{(2^{3})}^{n} + {(3^{3})}^{n}}{6^{n} (2^{n} + 3^{n})} = \frac{7}{6}$ $\frac{{(2^{n})}^{3} + {(3^{n})}^{3}}{6^{n} (2^{n} + 3^{n})} = \frac{7}{6}$ $\frac{(2^{n} + 3^{n}) (2^{2 n} - 2^{n} 3^{n} + 3^{2 n})}{6^{n} (2^{n} + 3^{n})} = \frac{7}{6}$ $6 . 2^{2 n} - 6 . 2^{n} 3^{n} + 6 . 3^{2 n} = 7 . 6^{n}$ $6 . 2^{2 n} - 13 . 2^{n} . 3^{n} + 6 . 3^{2 n} = 0$ $L e t 2^{n} = x & 3^{n} = y$ $6 x^{2} - 13 x y + 6 y^{2}$ $(3 x - 2 y) (2 x - 3 y) = 0$ $3 x = 2 y ∣ 2 x = 3 y$ $\frac{x}{y} = \frac{2}{3} ∣ \frac{x}{y} = \frac{3}{2}$ $\frac{2^{n}}{3^{n}} = {(\frac{2}{3})}^{1} ∣ \frac{2^{n}}{3^{n}} = {(\frac{2}{3})}^{- 1}$ ${(\frac{2}{3})}^{n} = {(\frac{2}{3})}^{1} ∣ {(\frac{2}{3})}^{n} = \frac{3}{2} = {(\frac{2}{3})}^{- 1}$ $n = 1 ∣ n = - 1$
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