How do you factor ${(3 x - 4)}^{3} + 27$ $(3x-4)^3 + 27$ ?

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How do you factor ${(3 x - 4)}^{3} + 27$ $(3x-4)^3 + 27$ ?

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Answer 1 · 2015-04-10T02:58:48

This is a sum of two cubes. Memorize the rule:

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

In this example, $a = (3 x - 4)$ $a=color(red)((3x-4))$ and $b = 3$ $b=3$ , so

${(3 x - 4)}^{3} + 3^{3} = ((3 x - 4) + 3) ({(3 x - 4)}^{2} - (3 x - 4) 3 + 3^{2})$ $color(red)((3x-4)^3)+3^3=(color(red)((3x-4))+3)(color(red)((3x-4)^2)-color(red)((3x-4))3+3^2)$

This answer can be simplified to get:

$(3 x - 1) (x^{2} - 24 x + 16 - 9 x + 12 + 9) = (3 x - 1) (x^{2} - 33 x + 37)$ $(3x-1)(x^2-24x+16-9x+12+9)=(3x-1)(x^2-33x+37)$

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How do you factor ${(3 x - 4)}^{3} + 27$ $(3x-4)^3 + 27$ ?

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