Q. Minimum of \(8^{x} + 8^{-x} – 4\left(4^{x} + 4^{-x}\right)\).

Put the expression in a simpler base. Let
\[ t = 2^{x}. \]
Then
\[ 4^{x} = (2^{2})^{x} = 2^{2x} = t^{2}, \qquad 8^{x} = (2^{3})^{x} = 2^{3x} = t^{3}, \]
and similarly for negative exponents
\[ 4^{-x} = t^{-2}, \qquad 8^{-x} = t^{-3}. \]
Since \(t=2^{x}>0\), we may work with \(t>0\).

Rewrite the function in terms of \(t\). The expression becomes
\[ f(x)=8^{x}+8^{-x}-4\bigl(4^{x}+4^{-x}\bigr)=t^{3}+t^{-3}-4\bigl(t^{2}+t^{-2}\bigr). \]

Introduce the symmetric variable
\[ y=t+t^{-1}. \]
Note that by AM-GM, \(y\ge 2\) for \(t>0\). Use the identities
\[ t^{2}+t^{-2}=y^{2}-2, \qquad t^{3}+t^{-3}=y^{3}-3y, \]
(the second follows from \((t+t^{-1})^{3}=t^{3}+t^{-3}+3(t+t^{-1})\) because \(tt^{-1}=1\)).

Express the function in terms of \(y\):
\[ f(x)=\bigl(y^{3}-3y\bigr)-4\bigl(y^{2}-2\bigr)=y^{3}-4y^{2}-3y+8. \]
Define \(g(y)=y^{3}-4y^{2}-3y+8\) with domain \(y\ge 2\).

Find critical points of \(g\). Compute the derivative:
\[ g'(y)=3y^{2}-8y-3. \]
Solve \(g'(y)=0\):
\[ 3y^{2}-8y-3=0 \implies y=\frac{8\pm\sqrt{64+36}}{6}=\frac{8\pm 10}{6}. \]
Thus the roots are
\[ y=3 \quad\text{and}\quad y=-\tfrac{1}{3}. \]
Only \(y=3\) lies in the domain \(y\ge 2\).

Evaluate \(g\) at the candidate and at the endpoint \(y=2\):
\[ g(3)=3^{3}-4\cdot 3^{2}-3\cdot 3+8=27-36-9+8=-10, \]
\[ g(2)=2^{3}-4\cdot 2^{2}-3\cdot 2+8=8-16-6+8=-6. \]
As \(y\to\infty\) the cubic term dominates and \(g(y)\to\infty\), so the minimum on \([2,\infty)\) occurs at \(y=3\).

Therefore the minimum value of the original expression is
\[ \min_{x\in\mathbb{R}}\bigl(8^{x}+8^{-x}-4(4^{x}+4^{-x})\bigr) = -10. \]

For completeness, find the \(x\) values where the minimum is attained. Solve
\[ t+t^{-1}=3 \]
which gives
\[ t^{2}-3t+1=0 \implies t=\frac{3\pm\sqrt{5}}{2}. \]
Since \(t=2^{x}\), the corresponding \(x\) values are
\[ x=\log_{2}\!\left(\frac{3+\sqrt{5}}{2}\right)\quad\text{and}\quad x=\log_{2}\!\left(\frac{3-\sqrt{5}}{2}\right). \]
Both give the same value \(-10\).