Solve x^6y^6-9x^3y^3+8

$$x ^ { 6 } y ^ { 6 } - 9 x ^ { 3 } y ^ { 3 } + 8$$

Factor

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

Solution Steps

Consider x^{6}y^{6}-9x^{3}y^{3}+8 as a polynomial over variable x.

$$y^{6}x^{6}-9y^{3}x^{3}+8$$

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

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

Consider x^{3}y^{3}-8. Rewrite x^{3}y^{3}-8 as \left(xy\right)^{3}-2^{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).

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

Consider x^{3}y^{3}-1. Rewrite x^{3}y^{3}-1 as \left(xy\right)^{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).

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

Rewrite the complete factored expression.

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

Evaluate

$$\left(xy\right)^{6}-9\left(xy\right)^{3}+8$$