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How do you simplify #(x-2)/(x^3-8)#?

Answer 1

Nathan Axel

#1/(x^2+2x+4)#

#x^3-8 = (x-2)(x^2+2x+4)# The above uses the rule #x^3-y^3 = (x-y)(x^2+xy+y^2)#

#(x-2)/(x^3-8) = (x-2)/((x-2)(x^2+2x+4)) = 1/(x^2+2x+4)#

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Answer 2

Eva Abramson

#1/(x^2+2x+4),x!=2#

#" "(x-2)/(x^3-8)#

In rational expressions, any variable on the denominator must not equal zero, so we can find restrictions from the original equation initially to save time in the end: #" "x^3-8!=0# #" "x^3!=8# #" "x!=2#

Continue solving by putting #8# into exponent form. #=(x-2)/(x^3-2^3)#

Factor the difference of cubes (#a^3-b^3=(a-b)(a^2+ab+b^2)#). #=(x-2)/((x-2)(x^2+2x+4))#

Reduce the fraction with #x-2#. #=cancel(x-2)/(cancel(x-2)(x^2+2x+4))# #=1/(x^2+2x+4),x!=2#

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Answer 3

Eva Abramson

To simplify the expression (x-2)/(x^3-8), we can factor the denominator as a difference of cubes. The denominator x^3-8 can be written as (x-2)(x^2+2x+4). Therefore, the expression simplifies to 1/(x^2+2x+4).

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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.