How do you factor $8 m^{3} - 125 n^{3}$ $8m^3 - 125n^3$ ?

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How do you factor $8 m^{3} - 125 n^{3}$ $8m^3 - 125n^3$ ?

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Answer 1 · 2018-05-16T20:24:50

$(2 m - 5 n) (4 m^{2} + 10 m n + 25 n^{2})$ $(2m-5n)(4m^2+10mn+25n^2)$

Explanation:

$this is a difference of cubes$ $"this is a "color(blue)"difference of cubes"$

$and factors in general as$ $"and factors in general as"$

$∙ x a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ $•color(white)(x)a^3-b^3=(a-b)(a^2+ab+b^2)$

$8 m^{3} = {(2 m)}^{3} \Rightarrow a = 2 m$ $8m^3=(2m)^3rArra=2m$

$125 n^{3} = {(5 n)}^{3} \Rightarrow b = 5 n$ $125n^3=(5n)^3rArrb=5n$

$8 m^{3} - 125 n^{3} = (2 m - 5 n) ({(2 m)}^{2} + (2 m \times 5 n) + {(5 n)}^{2})$ $8m^3-125n^3=(2m-5n)((2m)^2+(2mxx5n)+(5n)^2)$

$\times \times \times \times x = (2 m - 5 n) (4 m^{2} + 10 m n + 25 n^{2})$ $color(white)(xxxxxxxxx)=(2m-5n)(4m^2+10mn+25n^2)$

Answer 2 · 2018-05-16T20:27:03

$(2 m - 5 n) (4 m^{2} + 10 m n + 25 n^{2})$ $(2m-5n) (4m^2 + 10mn + 25n^2)$

Explanation:

Remember that $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ $a^3-b^3=(a-b)(a^2+ab+b^2)$

$8 m^{3} - 125 n^{3} = 2^{3} m^{3} - 5^{3} n^{3} = {(2 m)}^{3} - {(5 n)}^{3}$ $8m^3-125n^3=2^3m^3-5^3n^3=(2m)^3-(5n)^3$

Therefore, $a = 2 m, b = 5 n$ $a=2m, b=5n$

Sub in: $(2 m - 5 n) (4 m^{2} + 10 m n + 25 n^{2})$ $(2m-5n) (4m^2 + 10mn + 25n^2)$

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How do you factor $8 m^{3} - 125 n^{3}$ $8m^3 - 125n^3$ ?

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