Let the expression E = 8^a + 8^b - 3 cdot 2^{ | vedclass

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(D) The expression is $E = (2^a)^3 + (2^b)^3 - 3 \cdot 2^a \cdot 2^b$. Using the identity $x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)$,let

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$\sqrt{2}$

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(D) The expression is $E = (2^a)^3 + (2^b)^3 - 3 \cdot 2^a \cdot 2^b$. Using the identity $x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)$,let $x = 2^a$,$y = 2^b$,and $z = -1$. Then $E - 1 = (2^a)^3 + (2^b)^3 + (-1)^3 - 3(2^a)(2^b)(-1) = (2^a + 2^b - 1)((2^a)^2 + (2^b)^2 + 1 - 2^a 2^b + 2^a + 2^b)$. Alternatively,using $x^3 + y^3 + z^3 - 3xyz = \frac{1}{2}(x+y+z)((x-y)^2 + (y-z)^2 + (z-x)^2)$,for $E = (2^a)^3 + (2^b)^3 - 3 \cdot 2^a \cdot 2^b$,we can write $E = (2^a + 2^b + c)((2^a)^2 + (2^b)^2 + c^2 - 2^a 2^b - 2^a c - 2^b c) - c^3 + 3 \cdot 2^a 2^b c$. Setting $c = -1$,$E = (2^a + 2^b - 1)((2^a)^2 + (2^b)^2 + 1 - 2^a 2^b + 2^a + 2^b) + 1 - 3 \cdot 2^a 2^b$. Actually,$E = (2^a + 2^b)( (2^a)^2 - 2^a 2^b + (2^b)^2 ) - 3 \cdot 2^a 2^b$. For $a=b=0$,$E = 1 + 1 - 3 = -1$. Since $2^a, 2^b > 0$,the minimum value $p = -1$ occurs at $a = 0, b = 0$. Thus,$P = (0, 0)$ and $p = -1$. The line is $x + y + 2(-1) = 0$,which is $x + y - 2 = 0$. The perpendicular distance from $(0, 0)$ to $x + y - 2 = 0$ is $d = \frac{|0 + 0 - 2|}{\sqrt{1^2 + 1^2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.

Let the expression $E = 8^a + 8^b - 3 \cdot 2^{a+b}$ take its minimum value $p$ at $a = \alpha$ and $b = \beta$. Then,the perpendicular distance of the point $P(\alpha, \beta)$ from the line $x + y + 2p = 0$ is

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