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

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

Factor

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

Solution Steps

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

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

Factor out common term y^{3}-1 by using distributive property.

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

Consider x^{3}-1. Rewrite x^{3}-1 as x^{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).

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

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

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

Evaluate

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