Solve θ^6+B^3

$${ \theta }^{ 6 } + { B }^{ 3 }$$

Factor

$$\left(B+\theta ^{2}\right)\left(B^{2}-B\theta ^{2}+\theta ^{4}\right)$$

Solution Steps

Rewrite \theta ^{6}+B^{3} as B^{3}+\left(\theta ^{2}\right)^{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).

$$\left(B+\theta ^{2}\right)\left(B^{2}-B\theta ^{2}+\theta ^{4}\right)$$

Evaluate

$$B^{3}+\theta ^{6}$$