Solve y^6+7y^3-8

$$y ^ { 6 } + 7 y ^ { 3 } - 8$$

Factor

$$\left(y-1\right)\left(y+2\right)\left(y^{2}-2y+4\right)\left(y^{2}+y+1\right)$$

Solution Steps

Find one factor of the form y^{k}+m, where y^{k} divides the monomial with the highest power y^{6} and m divides the constant factor -8. One such factor is y^{3}+8. Factor the polynomial by dividing it by this factor.

$$\left(y^{3}+8\right)\left(y^{3}-1\right)$$

Consider y^{3}+8. Rewrite y^{3}+8 as y^{3}+2^{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).

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

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. The following polynomials are not factored since they do not have any rational roots: y^{2}+y+1,y^{2}-2y+4.

$$\left(y-1\right)\left(y^{2}+y+1\right)\left(y+2\right)\left(y^{2}-2y+4\right)$$

Evaluate

$$y^{6}+7y^{3}-8$$