How do you simplify # (x^3+27)/(9x+27) / (3x^2-9x+27)/(4x)#?

Question

How do you simplify # (x^3+27)/(9x+27)  /   (3x^2-9x+27)/(4x)#?

Isaiah Anthony · Accepted Answer

#(4x)/27#

Recall that division is the same as multiplying by the reciprocal. Thus, we can flip the right term and multiply instead of divide.

#=(x^3+27)/(9x+27)((4x)/(3x^2-9x+27))#

Factor each term.

The top left is a sum of cubes.

#=((x+3)(x^2-3x+9))/(9x+27)((4x)/(3x^2-9x+27))#

Factor a #9# from the bottom left.

#=((x+3)(x^2-3x+9))/(9(x+3))((4x)/(3x^2-9x+27))#

Cancel the #(x+3)# terms.

#=(x^2-3x+9)/(9)((4x)/(3x^2-9x+27))#

Factor a #3# from the bottom right.

#=(x^2-3x+9)/(9)((4x)/(3(x^2-3x+9)))#

Notice that the #x^2-3x+9# terms will cancel.

#=1/9((4x)/3)=(4x)/27#

Eva Adamson · Answer

undefined To simplify the expression (x^3+27)/(9x+27) / (3x^2-9x+27)/(4x), we can start by simplifying each fraction separately.

For the numerator (x^3+27), we can factor it using the sum of cubes formula: a^3 + b^3 = (a + b)(a^2 - ab + b^2). In this case, a = x and b = 3, so we have (x + 3)(x^2 - 3x + 9).

For the denominator (9x + 27), we can factor out the common factor of 9: 9(x + 3).

Next, let's simplify the second fraction (3x^2-9x+27)/(4x). We can factor out the common factor of 3: 3(x^2 - 3x + 9).

Now, we can rewrite the expression as (x + 3)(x^2 - 3x + 9) / 9(x + 3) divided by 3(x^2 - 3x + 9) / 4x.

To divide fractions, we can multiply the first fraction by the reciprocal of the second fraction. The reciprocal of 3(x^2 - 3x + 9) / 4x is 4x / 3(x^2 - 3x + 9).

Multiplying the fractions, we get (x + 3)(x^2 - 3x + 9) / 9(x + 3) * 4x / 3(x^2 - 3x + 9).

Now, we can cancel out the common factors in the numerator and denominator: (x + 3) / 9 * 4x / 3.

Multiplying the numerators and denominators, we have (4x(x + 3)) / (9 * 3).

Simplifying further, we get (4x^2 + 12x) / 27.

Therefore, the simplified expression is (4x^2 + 12x) / 27.