Solve 16x^4-8/9x^2y^2+1/81y^4

$$16 x ^ { 4 } - \frac { 8 } { 9 } x ^ { 2 } y ^ { 2 } + \frac { 1 } { 81 } y ^ { 4 }$$

Factor

$$\frac{\left(36x^{2}-y^{2}\right)^{2}}{81}$$

Solution Steps

Factor out \frac{1}{81}.

$$\frac{1296x^{4}-72x^{2}y^{2}+y^{4}}{81}$$

Consider 1296x^{4}-72x^{2}y^{2}+y^{4}. Consider 1296x^{4}-72x^{2}y^{2}+y^{4} as a polynomial over variable x.

$$1296x^{4}-72y^{2}x^{2}+y^{4}$$

Find one factor of the form kx^{m}+n, where kx^{m} divides the monomial with the highest power 1296x^{4} and n divides the constant factor y^{4}. One such factor is 36x^{2}-y^{2}. Factor the polynomial by dividing it by this factor.

$$\left(36x^{2}-y^{2}\right)\left(36x^{2}-y^{2}\right)$$

Consider 36x^{2}-y^{2}. Rewrite 36x^{2}-y^{2} as \left(6x\right)^{2}-y^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(6x-y\right)\left(6x+y\right)$$

Rewrite the complete factored expression.

$$\frac{\left(6x-y\right)^{2}\left(6x+y\right)^{2}}{81}$$

Evaluate

$$\frac{\left(36x^{2}-y^{2}\right)^{2}}{81}$$