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We know that $AM \geqslant GM$ for positive numbers.

$\large \displaystyle \therefore \frac{3^x + 3^{1-x}}{2} \geqslant \sqrt{3^x \times 3^{1-x}}$

$\large \displaystyle \implies \frac{3^x + 3^{1-x}}{2} \geqslant \sqrt{3^x \times \frac{3}{3^x}}$

$\large \displaystyle \implies 3^x + 3^{1-x} \geqslant 2\sqrt{3}$

The minimum value of the expression $\large \displaystyle 3^x + 3^{1-x} \text{ is } 2\sqrt{3}$ .

Where, $a = 2, \, b = 3$

$\large \displaystyle \therefore \color{#3D99F6}{a^3 + b^2} = \color{#20A900}{2^3 + 3^2} = \color{#D61F06}{8+9} = \color{#69047E}{\boxed{17}}.$

let f(x)= 3^x + 3^(1-x) the minimum value occurs when f'(x)=0 then

3^x * log3 - 3^(1-x) * log3 = 0 then

3^x - 3^(1-x) = 0

3^x = 3^(1-x) then

x = 1-x

2x = 1

x = 1/2 the minimum value occurs at x = 1/2 then

the minimum value of 3^x + 3^(1-x) is 3^(1/2) + 3^(1-1/2)

which is 3^(1/2) + 3^(1/2) = 2 * sqrt(3)

then a = 2 and b = 3

and a^3 + b^2 = 2^3 + 3^2

          = 8 + 9

      =17