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Difficulty: HardEquivalent Algebraic Expressions

For all x>0x > 0, which of the following is equivalent to the expression (x−1x)2(x+1+1x)\left(\sqrt{x} - \frac{1}{\sqrt{x}}\right)^2 \left(x + 1 + \frac{1}{x}\right)?

  1. A
    (x+1)(x3−1)x2\frac{(x+1)(x^3-1)}{x^2}
  2. B
    (x2+1)(x2+x+1)x2\frac{(x^2+1)(x^2+x+1)}{x^2}
  3. (x−1)(x3−1)x2\frac{(x-1)(x^3-1)}{x^2}Answer
  4. D
    (x−1)(x3+1)x2\frac{(x-1)(x^3+1)}{x^2}

Answer

(x−1)(x3−1)x2\frac{(x-1)(x^3-1)}{x^2}
The correct expression is obtained by first converting x−1x\sqrt{x} - \frac{1}{\sqrt{x}} to x−1x\frac{x-1}{\sqrt{x}}, and then squaring it to get (x−1)2x\frac{(x-1)^2}{x}. Next, the term x+1+1xx + 1 + \frac{1}{x} is converted to x2+x+1x\frac{x^2+x+1}{x}. Multiplying these two results yields (x−1)2(x2+x+1)x2\frac{(x-1)^2(x^2+x+1)}{x^2}. Grouping (x−1)(x2+x+1)(x-1)(x^2+x+1) and substituting it with the difference of cubes identity, x3−1x^3-1, gives the final equivalent expression (x−1)(x3−1)x2\frac{(x-1)(x^3-1)}{x^2}.

Step-by-Step Solution

1
Simplify the first factor by writing x−1x\sqrt{x} - \frac{1}{\sqrt{x}} under a common denominator and squaring it.
(x−1x)2=(x−1)2x\left(\frac{x-1}{\sqrt{x}}\right)^2 = \frac{(x-1)^2}{x}
This allows us to combine the terms in the first factor before multiplication.
2
Write the second factor x+1+1xx + 1 + \frac{1}{x} under a common denominator.
x2+x+1x\frac{x^2+x+1}{x}
This allows us to multiply the two fractional factors easily.
3
Multiply the two simplified factors together.
(x−1)2(x2+x+1)x2=(x−1)(x−1)(x2+x+1)x2\frac{(x-1)^2(x^2+x+1)}{x^2} = \frac{(x-1)(x-1)(x^2+x+1)}{x^2}
Multiplying the numerators and denominators of the fractions gives the product of the terms.
4
Apply the difference of cubes factorization formula: (x−1)(x2+x+1)=x3−1(x-1)(x^2+x+1) = x^3-1.
(x−1)(x3−1)x2\frac{(x-1)(x^3-1)}{x^2}
This simplifies the product to its final equivalent form.

Key Concept

Simplifying rational expressions and applying algebraic identities (specifically the binomial square and difference of cubes).

Alternative Method

Alternatively, you can substitute a convenient value for xx, such as x=4x = 4, into the original expression and each of the options, then compare the results.
Estimated Time:2m 0s
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Topics

Equivalent Algebraic Expressions

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