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How do you factor #125x^3+8g^3#?

Answer 1

Isaiah Anthony

#125x^3+8g^3=(5x+2g)(25x^2-10xg+4g^2)#

As #125x^3+8g^3=(5x)^3+(2g)^3#, it can be factorized using identity #x^3+y^3=(x+y)(x^2-xy+y^2)#

and #125x^3+8g^3=(5x)^3+(2g)^3#

= #(5x+2g)((5x)^2-(5x)xx(2g)+(2g)^2)#

= #(5x+2g)(25x^2-10xg+4g^2)#

For a proof of the identity see below.

#x^3+y^3#

= #x^3+x^2y-x^2y-xy^2+xy^2+y^3#

= #x^2(x+y)-xy(x+y)+y^2(x+y)#

= #(x+y)(x^2-xy+y^2)#

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Answer 2

Jameson Allen

To factor the expression (125x^3 + 8g^3), you can use the sum of cubes formula, which states that (a^3 + b^3 = (a + b)(a^2 - ab + b^2)). Applying this formula, we have:

[125x^3 + 8g^3 = (5x)^3 + (2g)^3]

So, using the formula, we get:

[= (5x + 2g)((5x)^2 - (5x)(2g) + (2g)^2)]

[= (5x + 2g)(25x^2 - 10xg + 4g^2)]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.