How do you factor # 27g^3 + 343#?

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Answer 1 · 2016-03-11T16:27:52

#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# is a perfect cube.

#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)#