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

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Answer 1 · 2016-04-11T19:26:53

$49x^6+126x^3y^2+81y^4 = (7x^3+9y^2)^2$

This is a perfect square trinomial of the form:

$a^2+2ab+b^2 = (a+b)^2$

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

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

$= (7x^3)^2+2(7x^3)(9y^2)+(9y)^2$

$= (7x^3+9y^2)^2$

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

How do you factor the expression 49x^6 + 126x^3y^2 + 81y^449x^6 + 126x^3y^2 + 81y^4?