How do you factor $128 x^{3} - 1024$ $128x^3 - 1024$ ?

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How do you factor $128 x^{3} - 1024$ $128x^3 - 1024$ ?

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Answer 1 · 2017-02-06T20:27:05

$128 (x - 2) (x^{2} + 2 x + 4)$ $128(x-2)(x^2+2x+4)$

Explanation:

The 2 terms have a $common factor$ $color(blue)"common factor"$ of 128, which can be taken out.

$\Rightarrow 128 (x^{3} - 8) \to (A)$ $rArr128(x^3-8)to(A)$

$x^{3} - 8$ $x^3-8$ is a $difference of cubes$ $color(blue)"difference of cubes"$ which, in general, is factorised as shown.

$color(red)(bar(ul(|color(white)(2/2)color(black)(a^3-b^3=(a-b)(a^2+ab+b^2))color(white)(2/2)|)))$

using a = x and b = 2, then

$rArrx^3-8=(x-2)(x^2+2x+2^2)=(x-2)(x^2+2x+4)$

Returning to (A) the complete factorisation is.

$128x^3-1024=128(x-2)(x^2+2x+4)$

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How do you factor $128 x^{3} - 1024$ $128x^3 - 1024$ ?

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How do you factor $128 x^{3} - 1024$ $128x^3 - 1024$ ?