How do you factor $5 x^{3} + 40 y^{6}$ $5x^3 + 40y^6$ ?

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How do you factor $5 x^{3} + 40 y^{6}$ $5x^3 + 40y^6$ ?

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Answer 1 · 2015-05-18T21:35:17

You can use a formula for a sum of cubes:
$a^{3} + b^{3} = (a + b) \cdot (a^{2} - a b + b^{2})$ $a^3+b^3=(a+b)*(a^2-ab+b^2)$

Notice that $40 y^{6} = 5 \cdot 8 \cdot y^{6} = 5 \cdot 2^{3} \cdot {(y^{2})}^{3} = 5 \cdot {(2 y^{2})}^{3}$ $40y^6=5*8*y^6=5*2^3*(y^2)^3=5*(2y^2)^3$

Using the above formula for $a = x$ $a=x$ and $b = 2 y^{2}$ $b=2y^2$ and factoring out constant $5$ $5$ , we get:
$5 x^{3} + 40 y^{6} = 5 x^{3} + 5 {(2 y^{2})}^{3} = 5 (x + 2 y^{2}) (x^{2} - 2 x y^{2} + 4 y^{4}))$ $5x^3+40y^6 =5x^3+5(2y^2)^3=5(x+2y^2)(x^2-2xy^2+4y^4))$

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How do you factor $5 x^{3} + 40 y^{6}$ $5x^3 + 40y^6$ ?

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