# Step-by-step Solution

## Simplify the quotient of powers $\frac{x^5\left(x-1\right)^3}{\left(x+2\right)^3\left(x^2+1\right)^3}$

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### Videos

$\frac{x^5\left(x^3-3x^2+3x-1\right)}{\left(x+2\right)^3\left(x^2+1\right)^3}$

## Step-by-step Solution

Problem to solve:

$\frac{x^5\left(x-1\right)^3}{\left(x+2\right)^3\left(x^2+1\right)^3}$

Solving method

1

The cube of a binomial (difference) is equal to the cube of the first term, minus three times the square of the first by the second, plus three times the first by the square of the second, minus the cube of the second term. In other words: $(a-b)^3=a^3-3a^2b+3ab^2-b^3 = (x)^3+3(x)^2(-1)+3(x)(-1)^2+(-1)^3 =$

$\frac{x^5\left(x^3-3x^2+3x-1\right)}{\left(x+2\right)^3\left(x^2+1\right)^3}$

$\frac{x^5\left(x^3-3x^2+3x-1\right)}{\left(x+2\right)^3\left(x^2+1\right)^3}$
$\frac{x^5\left(x-1\right)^3}{\left(x+2\right)^3\left(x^2+1\right)^3}$