Solve -192+3r^3

$$- 192 + 3 r ^ { 3 }$$

Factor

$$3\left(r-4\right)\left(r^{2}+4r+16\right)$$

Solution Steps

Factor out 3.

$$3\left(-64+r^{3}\right)$$

Consider -64+r^{3}. Rewrite -64+r^{3} as r^{3}-4^{3}. The difference of cubes can be factored using the rule: a^{3}-b^{3}=\left(a-b\right)\left(a^{2}+ab+b^{2}\right).

$$\left(r-4\right)\left(r^{2}+4r+16\right)$$

Rewrite the complete factored expression. Polynomial r^{2}+4r+16 is not factored since it does not have any rational roots.

$$3\left(r-4\right)\left(r^{2}+4r+16\right)$$

Evaluate

$$3\left(r^{3}-64\right)$$