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and evaluating   it in  $(1, 1, 1)$ we get $f (1, 1, 1) = 1 \cdot 1 + 1 \cdot 1 = 1 + 1 = 0$. \\  Note that the inputs input  of $f$ is a $3$-tuple of bits and its output is only one bit, i.e., $0$ or $1$.\\ bit.\\  Let us now consider also  another function $g:(\Fb)^3 \rightarrow \Fb$ defined by \Fb$,  $g(x, y, z) = x^2 y + yz$ \\  Wewant to  evaluate $g$ at the same points $3$-tuples  $(0, 0, 0)$ and $(1, 1, 1)$ as   before; and  we obtain that  $g(0, 0, 0) 0)=f(0,0,0)  = 0$ and $g(1, 1, 1) =f(1, 1, 1)  = 0$. We Indeed, you  can check that $f$ and $g$ have take  the same values in \emph{all} points vectors  of $(\Fb)^3$ (and we leave this computation to the reader), $(\Fb)^3$,  so we have that  $$f (x, y, z) = g(x, y, z), \; \forall (x, y, z) \in (\Fb)^3$$  Hence (\Fb)^3.$$  Hence,  $f$ and $g$ are distinct \textbf{distinct}  as polynomials polynomials,  but they  are equal \textbf{equal}  as functions, since given an input,  they give return  the same output for each input. output!  \\  \smallskip  Since $x^2 = x \forall x \in \Fb $, then every polynomial function   from $(\Fb)^3$  to $\Fb$ can be written as a polynomial where the degree of