Solve x^5-x^3+27x^2-27=

$$x ^ { 5 } - x ^ { 3 } + 27 x ^ { 2 } - 27 =$$

Factor

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

Solution Steps

Do the grouping x^{5}-x^{3}+27x^{2}-27=\left(x^{5}-x^{3}\right)+\left(27x^{2}-27\right), and factor out x^{3} in the first and 27 in the second group.

$$x^{3}\left(x^{2}-1\right)+27\left(x^{2}-1\right)$$

Factor out common term x^{2}-1 by using distributive property.

$$\left(x^{2}-1\right)\left(x^{3}+27\right)$$

Consider x^{2}-1. Rewrite x^{2}-1 as x^{2}-1^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-1\right)\left(x+1\right)$$

Consider x^{3}+27. Rewrite x^{3}+27 as x^{3}+3^{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).

$$\left(x+3\right)\left(x^{2}-3x+9\right)$$

Rewrite the complete factored expression. Polynomial x^{2}-3x+9 is not factored since it does not have any rational roots.

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

Evaluate

$$x^{5}-x^{3}+27x^{2}-27$$