If 9^x +9^(−x) = 3^(2+x) +3^(2−x) −20, then 27^x +27^(−x) =?


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Question Number 115018 by bemath last updated on 23/Sep/20

$I f 9^{x} + 9^{- x} = 3^{2 + x} + 3^{2 - x} - 20, t h e n$ $27^{x} + 27^{- x} = ?$
	Answered by bobhans last updated on 23/Sep/20
	$⇒ (3^x )^2 +(3^(−x) )^2 =9.3^x +9.3^(−x) −20 ⇒ set 3^x =a ∧3^(−x) =b ⇒(a+b)^2 −2=9(a+b)−20 ⇒(a+b)^2 −9(a+b)+18=0 ⇒ a+b = 3 ∨ a+b = 6 now 27^x +27^(−x) =a^3 +b^3 =(a+b)^3 −3ab(a+b) where ab = 1 ⇔ a^3 +b^3 =(a+b){(a+b)^2 −3} for a+b = 3 ⇒ a^3 +b^3 = 3×6=18 for a+b = 6 ⇒a^3 +b^3 = 6×33=198$
	$\Rightarrow {(3^{x})}^{2} + {(3^{- x})}^{2} = 9 . 3^{x} + 9 . 3^{- x} - 20$ $\Rightarrow s e t 3^{x} = a \land 3^{- x} = b$ $\Rightarrow {(a + b)}^{2} - 2 = 9 (a + b) - 20$ $\Rightarrow {(a + b)}^{2} - 9 (a + b) + 18 = 0$ $\Rightarrow a + b = 3 \lor a + b = 6$ $n o w 27^{x} + 27^{- x} = a^{3} + b^{3} = {(a + b)}^{3} - 3 a b (a + b)$ $w h e r e a b = 1$ $\Leftrightarrow a^{3} + b^{3} = (a + b) {{(a + b)}^{2} - 3}$ $f o r a + b = 3 \Rightarrow a^{3} + b^{3} = 3 \times 6 = 18$ $f o r a + b = 6 \Rightarrow a^{3} + b^{3} = 6 \times 33 = 198$
	Answered by PRITHWISH SEN 2 last updated on 23/Sep/20
	$let 3^x =a a^2 +(1/a^2 ) = 9(a+(1/a))−20 t^2 −9t+18=0 {let t=(a+(1/a))} t=3,6 27^x +27^(−x) = (a^3 +(1/a^3 ))= 3^3 −3.3=18 when t=3 = 6^3 −3.6=216−18=198$
	$let 3^{x} = a$ $a^{2} + \frac{1}{a^{2}} = 9 (a + \frac{1}{a}) - 20$ $t^{2} - 9 t + 18 = 0 {let t = (a + \frac{1}{a})}$ $t = 3, 6$ $27^{x} + 27^{- x} = (a^{3} + \frac{1}{a^{3}}) = 3^{3} - 3.3 = 18 when t = 3$ $= 6^{3} - 3.6 = 216 - 18 = 198$
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