How do you factor $8 x^{3} \cdot y^{6} + 27$ $8x^3*y^6 + 27$ ?

Question

2001 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-13T23:57:21

$(2 x y^{2} + 3) (4 x^{2} y^{4} - 6 x y^{2} + 9)$ $(2xy^2+3)(4x^2y^4-6xy^2+9)$

Sum of cubes: $(a^{3} + b^{3}) = (a + b) (a^{2} - a b + b^{2})$ $(a^3+b^3)=(a+b)(a^2-ab+b^2)$

Since $8 x^{3} y^{6} + 27 = {(2 x y^{2})}^{3} + {(3)}^{3}$ $8x^3y^6+27=(2xy^2)^3+(3)^3$ ,

$a = 2 x y^{2}$ $a=2xy^2$
$b = 3$ $b=3$

$8 x^{3} y^{6} + 27 = (2 x y^{2} + 3) (4 x^{2} y^{4} - 6 x y^{2} + 9)$ $8x^3y^6+27=(2xy^2+3)(4x^2y^4-6xy^2+9)$

How do you factor 8x^3*y^6 + 278x3⋅y6+278x^3*y^6 + 27?