How do you factor $m^{3} + {(m + 3)}^{3}$ $m^3 + (m+3)^3$ ?

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How do you factor $m^{3} + {(m + 3)}^{3}$ $m^3 + (m+3)^3$ ?

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Answer 1 · 2018-05-21T09:45:41

$(2 m + 3) \cdot (m^{2} + 3 m + 9)$ $(2m+3)*(m^2+3m+9)$

Explanation:

Using ${(a + b)}^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}$ $(a+b)^3=a^3+3a^2b+3ab^2+b^3$

Answer 2 · 2018-05-21T10:07:51

$(2 m + 3) (m^{2} + 3 m + 9)$ $(2m+3)(m^2+3m+9)$

Explanation:

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

$which factors in general as$ $"which factors in general as"$

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

$here a = m and b = m + 3$ $"here "a=m" and "b=m+3$

$\Rightarrow m^{3} + {(m + 3)}^{3}$ $rArrm^3+(m+3)^3$

$= (m + m + 3) (m^{2} - m (m + 3) + {(m + 3)}^{2})$ $=(m+m+3)(m^2-m(m+3)+(m+3)^2)$

$= (2 m + 3) (m^{2} - m^{2} - 3 m + m^{2} + 6 m + 9)$ $=(2m+3)(m^2-m^2-3m+m^2+6m+9)$

$= (2 m + 3) (m^{2} + 3 m + 9)$ $=(2m+3)(m^2+3m+9)$

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How do you factor $m^{3} + {(m + 3)}^{3}$ $m^3 + (m+3)^3$ ?

2 Answers

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