Solve a^{2}bx^{6}y^{2}-c^{3}z^{4}-a^{2}bz^4+c^3x^6y^2

$$a ^ { 2 } b x ^ { 6 } y ^ { 2 } - c ^ { 3 } z ^ { 4 } - a ^ { 2 } b z ^ { 4 } + c ^ { 3 } x ^ { 6 } y ^ { 2 }$$

Factor

$$\left(ba^{2}+c^{3}\right)\left(yx^{3}-z^{2}\right)\left(yx^{3}+z^{2}\right)$$

Solution Steps

Do the grouping a^{2}bx^{6}y^{2}-c^{3}z^{4}-a^{2}bz^{4}+c^{3}x^{6}y^{2}=\left(a^{2}bx^{6}y^{2}-a^{2}bz^{4}\right)+\left(c^{3}x^{6}y^{2}-c^{3}z^{4}\right), and factor out ba^{2} in the first and c^{3} in the second group.

$$ba^{2}\left(y^{2}x^{6}-z^{4}\right)+c^{3}\left(y^{2}x^{6}-z^{4}\right)$$

Factor out common term y^{2}x^{6}-z^{4} by using distributive property.

$$\left(y^{2}x^{6}-z^{4}\right)\left(ba^{2}+c^{3}\right)$$

Consider y^{2}x^{6}-z^{4}. Rewrite y^{2}x^{6}-z^{4} as \left(x^{3}y\right)^{2}-\left(z^{2}\right)^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

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

Reorder the terms.

$$\left(yx^{3}-z^{2}\right)\left(yx^{3}+z^{2}\right)$$

Rewrite the complete factored expression.

$$\left(ba^{2}+c^{3}\right)\left(yx^{3}-z^{2}\right)\left(yx^{3}+z^{2}\right)$$

Evaluate

$$\left(ba^{2}+c^{3}\right)\left(y^{2}x^{6}-z^{4}\right)$$