Question

Dacă 3^a+3^(-a)=4, atunci 27^a+27^(-a)=?

Answer 1

$3^a + 3^{-a} = 4 \\ \\ 3^a + 3^{-a} = 4 \Big|^2 \Rightarrow( 3^a + 3^{-a} )^2= 4^2 \Rightarrow (3^a)^2 + 2\cdot 3^a\cdot 3^{-a} + (3^{-a})^2 = 16 \Rightarrow \\ \\ \Rightarrow (3^a)^2 + 2 + (3^{-a})^2 = 16 \Rightarrow (3^a)^2 + (3^{-a})^2 = 16-2 \Rightarrow 3^a)^2 + (3^{-a})^2 = 14\\ \\ \\ 27^a + 27^{-a} = (3^3)^a + (3^3)^{-a} = (3^a)^3 + (3^{-a})^3= \\ \\ = \Big(3^a+ 3^{-a} \Big)\Big((3^a)^2 - 3^a\cdot 3^{-a} + (3^{-a})^2\Big) = \\ \\ = 4 \Big((3^a)^2 - 3^{a-a} + (3^{-a})^2\Big) =\\ \\ = 4 \Big((3^a)^2 - 3^{0} + (3^{-a})^2\Big) = \\ \\= 4 \Big((3^a)^2 - 1 + (3^{-a})^2\Big) = 4(14-1) = 4\cdot 13 = 52$