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Chapter 2: Problem 91

Perform the division by assuming that \(n\) is a positive integer. \(\frac{x^{3 n}+9 x^{2 n}+27 x^{n}+27}{x^{n}+3}\)

Short Answer

Expert verified
The division \(\frac{x^{3 n}+9 x^{2 n}+27 x^{n}+27}{x^{n}+3}\) simplifies to \((x^n + 3)^2\).

Step by step solution

01

Recognize Expression Pattern

First, we observe and recognize the pattern in the given expression. The form of expression \(\frac{x^{3n} + 9x^{2n} + 27x^n + 27}{x^n + 3}\) is very reminiscent of both the cube of a binomial \((a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\) and the sum of cubes \(a^3 + b^3 = (a+b)(a^2 - ab + b^2)\) pattern.
02

Apply the binomial theorem

Taking the numerator \(x^{3n} + 9x^{2n} + 27x^n + 27\), it can be rewritten as \((x^n)^3 + 3*3(x^n)^2 + 3*3^2x^n + 3^3\), which fits the identity of \((a + b)^3\). Our 'a' is \(x^n\) and 'b' is 3. Therefore, we can write the numerator as \((x^n + 3)^3\).
03

Simplify by the denominator

Now, we have simplified our expression to \(\frac{(x^n + 3)^3}{x^n + 3}\). We cancel out the \(x^n + 3\) in the numerator and the denominator and we are left with \((x^n + 3)^2\).

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Solve the inequality and graph the solution on the real number line. \(\frac{x^{2}+x-6}{x} \geq 0\)

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