Solve y^4-3y^3+y-3

$$y ^ { 4 } - 3 y ^ { 3 } + y - 3$$

Factor

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

Solution Steps

Do the grouping y^{4}-3y^{3}+y-3=\left(y^{4}-3y^{3}\right)+\left(y-3\right), and factor out y^{3} in y^{4}-3y^{3}.

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

Factor out common term y-3 by using distributive property.

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

Consider y^{3}+1. Rewrite y^{3}+1 as y^{3}+1^{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).

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

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

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

Evaluate

$$y^{4}-3y^{3}+y-3$$