How do you factor $x^{6} + 2 x^{3} y^{2} + y^{4}$ $x^6 + 2x^3y^2 + y^4$ ?

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How do you factor $x^{6} + 2 x^{3} y^{2} + y^{4}$ $x^6 + 2x^3y^2 + y^4$ ?

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Answer 1 · 2016-01-03T15:07:43

$x^{6} + 2 x^{3} y^{2} + y^{4} = {(x^{3} + y^{2})}^{2}$ $x^6+2x^3y^2+y^4 = (x^3+y^2)^2$

Explanation:

We can recognise this as a perfect square trinomial, since it is of the form $a^{2} + 2 a b + b^{2}$ $a^2+2ab+b^2$ with $a = x^{3}$ $a=x^3$ and $b = y^{2}$ $b=y^2$ .

$x^{6} + 2 x^{3} y^{2} + y^{4}$ $x^6+2x^3y^2+y^4$

$= {(x^{3})}^{2} + 2 (x^{3}) (y^{2}) + {(y^{2})}^{2}$ $=(x^3)^2 + 2(x^3)(y^2) + (y^2)^2$

$= {(x^{3} + y^{2})}^{2}$ $=(x^3+y^2)^2$

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How do you factor $x^{6} + 2 x^{3} y^{2} + y^{4}$ $x^6 + 2x^3y^2 + y^4$ ?

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Explanation:

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