Solve 64/27+y^3=

$$\frac { 64 } { 27 } + y ^ { 3 } =$$

Factor

$$\frac{\left(3y+4\right)\left(9y^{2}-12y+16\right)}{27}$$

Solution Steps

Factor out \frac{1}{27}.

$$\frac{64+27y^{3}}{27}$$

Consider 64+27y^{3}. Rewrite 64+27y^{3} as \left(3y\right)^{3}+4^{3}. The sum of cubes can be factored using the rule: a^{3}+b^{3}=\left(a+b\right)\left(a^{2}-ab+b^{2}\right).

$$\left(3y+4\right)\left(9y^{2}-12y+16\right)$$

Rewrite the complete factored expression. Polynomial 9y^{2}-12y+16 is not factored since it does not have any rational roots.

$$\frac{\left(3y+4\right)\left(9y^{2}-12y+16\right)}{27}$$

Evaluate

$$y^{3}+\frac{64}{27}$$