If-9-x-9-x-3-2-x-3-2-x-20-then-27-x-27-x-

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

\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} + \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