How do you factor $27 g^{3} + 343$ $27g^3 + 343$ ?

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Answer 1 · 2016-03-11T16:27:52

$27 g^{3} + 343 = (3 g + 7) (9 g^{2} - 21 g + 49)$ $27g^3+343=(3g+7)(9g^2-21g+49)$

We need to write the expression in a form that remembers the remarkable identities.

In this case we could try to see if $343$ $343$ is a perfect cube.

$343 = 7^{3}$ $343=7^3$

$:. 27g^3+343=(3g)^3+7^3$

Now it's a sum of cubes then we could use the following identity:

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

$:.(3g)^3+7^3=(3g+7)((3g)^2-3g*7+7^2)=$
$=(3g+7)(9g^2-21g+49)$

How do you factor 27g^3 + 34327g3+343 27g^3 + 343?