$${ x }^{ 3 } - \frac{ 1 }{ 4 } x$$
$\frac{x\left(2x-1\right)\left(2x+1\right)}{4}$
$$\frac{4x^{3}-x}{4}$$
$$x\left(4x^{2}-1\right)$$
$$\left(2x-1\right)\left(2x+1\right)$$
$$\frac{x\left(2x-1\right)\left(2x+1\right)}{4}$$
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$x^{3}-\frac{x}{4}$