Solve 3a^11-9a^8+9a^5-3a^2

$$3 a ^ { 11 } - 9 a ^ { 8 } + 9 a ^ { 5 } - 3 a ^ { 2 }$$

Factor

$$3a^{2}\left(a-1\right)^{3}\left(a^{2}+a+1\right)^{3}$$

Solution Steps

Factor out 3.

$$3\left(a^{11}-3a^{8}+3a^{5}-a^{2}\right)$$

Consider a^{11}-3a^{8}+3a^{5}-a^{2}. Factor out a^{2}.

$$a^{2}\left(a^{9}-3a^{6}+3a^{3}-1\right)$$

Consider a^{9}-3a^{6}+3a^{3}-1. Find one factor of the form a^{k}+m, where a^{k} divides the monomial with the highest power a^{9} and m divides the constant factor -1. One such factor is a^{3}-1. Factor the polynomial by dividing it by this factor.

$$\left(a^{3}-1\right)\left(a^{6}-2a^{3}+1\right)$$

Consider a^{3}-1. Rewrite a^{3}-1 as a^{3}-1^{3}. The difference of cubes can be factored using the rule: p^{3}-q^{3}=\left(p-q\right)\left(p^{2}+pq+q^{2}\right).

$$\left(a-1\right)\left(a^{2}+a+1\right)$$

Consider a^{6}-2a^{3}+1. Find one factor of the form a^{n}+u, where a^{n} divides the monomial with the highest power a^{6} and u divides the constant factor 1. One such factor is a^{3}-1. Factor the polynomial by dividing it by this factor.

$$\left(a^{3}-1\right)\left(a^{3}-1\right)$$

Consider a^{3}-1. Rewrite a^{3}-1 as a^{3}-1^{3}. The difference of cubes can be factored using the rule: p^{3}-q^{3}=\left(p-q\right)\left(p^{2}+pq+q^{2}\right).

$$\left(a-1\right)\left(a^{2}+a+1\right)$$

Rewrite the complete factored expression. Polynomial a^{2}+a+1 is not factored since it does not have any rational roots.

$$3a^{2}\left(a-1\right)^{3}\left(a^{2}+a+1\right)^{3}$$

Evaluate

$$3a^{2}\left(a^{3}-1\right)^{3}$$