Solve y^4+y^3-y-1=

$$y ^ { 4 } + y ^ { 3 } - y - 1 =$$

Factor

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

Solution Steps

Do the grouping y^{4}+y^{3}-y-1=\left(y^{4}+y^{3}\right)+\left(-y-1\right), and factor out y^{3} in the first and -1 in the second group.

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

Factor out common term y+1 by using distributive property.

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

Consider y^{3}-1. Rewrite y^{3}-1 as y^{3}-1^{3}. The difference 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.

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

Evaluate

$$y^{4}+y^{3}-y-1$$