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Scale the equal intervals to unit length. A nonzero compactly supported cubic B-spline with maximum continuity is $C^2$, so its fourth derivative is a sum of impulses at the knots. Equivalently it is a combination of truncated cubes,To vanish to the right of its support, the polynomial tail must be identically zero. Expanding that tail requires ${\sum }_j{a}_j{j}^r=0$ for $r=0,1,2,3$. Four or fewer distinct knots give a full-rank Vandermonde matrix and only the zero solution. Hence a nonzero function requires at least five knots, or four intervals. At knots $0,1,2,3,4$, the one-dimensional nullspace has coefficients proportional to $(1,-4,6,-4,1)$. The unscaled combination integrates to six on $[0,4]$, so unit-integral cardinal normalization gives:This is the Cardinal cubic B-spline, nonnegative with support $[0,4]$. To obtain its cubic B-spline Bézier extraction, use local parameter $t=x-j\in [0,1]$ on interval $[j,j+1]$. The four power-form numerators, with a common divisor six, areFor $a+bt+c{t}^2+d{t}^3$, the cubic Bernstein basis coefficients are $a$, $a+b/3$, $a+2b/3+c/3$, $a+b+c+d$. Applying this conversion givesEach scalar row describes a cubic Bézier curve graph piece with abscissa controls $j,j+1/3,j+2/3,j+1$. The truncated-cube construction proves value, first- and second-derivative agreement at all joins, including the zero exterior. Minimum support and maximum continuity determine the basis up to a scalar; the displayed normalization fixes that remaining freedom.