Volume V = 576 cm³, height h = 4 * width w, length l = 6 * height h

Express all dimensions in terms of $w$: $h = 4w$ and $l = 6(4w) = 24w$. Thus, $V = 24w \times w \times 4w = 96w^3$. Since $96 \times 6 = 576$, $w^3 = 6$ is wrong, check: $96 \times 2^3 = 96 \times 8 = 768$ (Wait, check math: $24 \times 4 = 96$. $576 / 96 = 6$. $w^3 = 6$? Re-read: '6 times as long as it is high'. $h=4w$, $l=6h=24w$. $V=w(4w)(24w)=96w^3$. $96w^3=576 \implies w^3=6$. Wait, $2^3=8$. Check options: $w=2 \implies V=96(8)=768$. Let's re-calculate: If $w=2$, $h=8$, $l=48$. $V=2 \times 8 \times 48 = 768$. If $V=576$, $w^3 = 576/96 = 6$. The cube root of 6 is not an integer. Let's re-read: '4 times as high as wide' ($h=4w$), '6 times long as wide'? No, '6 times as long as it is high' ($l=6h=24w$). Maybe '6 times as long as it is wide'? $l=6w$. $V=w(4w)(6w)=24w^3$. $24w^3=576 \implies w^3=24$. Still not matching. Let's re-calculate: $V = w \times (4w) \times (6w)$ is not the text. Text: $h=4w$, $l=6h=24w$. Maybe $V=w(4w)(6)=576 \implies 24w^2=576 \implies w^2=24$. Check options again.

Confusing the relationships between length, width, and height by failing to express them all in terms of a single variable.

Always verify units (cm³) and express complex geometric relationships as a single-variable algebraic equation before solving.