$\frac{e^{3x}+e^{-3x}}{e^{x}+e^{-x}}$

we get value of x by taking log to the base e on both sides to be ln (sqrt 7) when put this value in the eqn and using the property that e^ln(a) = a we get a simplified ans 43/7

$\dfrac{e^{3x}+e^{-3x}}{e^{x}+e^{-x}}$

$=\dfrac{(e^x+e^{-x})(e^{2x}-e^0+e^{-2x})}{e^{x}+e^{-x}}$

$=e^{2x}-e^0+e^{-2x}$

$=7-1+\dfrac{1}{7}$

$=\dfrac{43}{7}$

Note that $\boxed{a^3+b^3=(a+b)(a^2-ab+b^2)}$

Where $a=e^x,b=e^{-x}$