# How do you factor the expression 49x^6 + 126x^3y^2 + 81y^4?

Apr 11, 2016

$49 {x}^{6} + 126 {x}^{3} {y}^{2} + 81 {y}^{4} = {\left(7 {x}^{3} + 9 {y}^{2}\right)}^{2}$

#### Explanation:

This is a perfect square trinomial of the form:

${a}^{2} + 2 a b + {b}^{2} = {\left(a + b\right)}^{2}$

with $a = 7 {x}^{3}$ and $b = 9 {y}^{2}$

$49 {x}^{6} + 126 {x}^{3} {y}^{2} + 81 {y}^{4}$

$= {\left(7 {x}^{3}\right)}^{2} + 2 \left(7 {x}^{3}\right) \left(9 {y}^{2}\right) + {\left(9 y\right)}^{2}$

$= {\left(7 {x}^{3} + 9 {y}^{2}\right)}^{2}$

No further factorisation is possible since the remaining terms in $x$ and $y$ are of distinct prime degrees.