The simplified form of the expression $\frac{\left(x^2-9\right) ×\left(9 x^2-1\right)}{\left(x-\frac{1}{3}\right) ×\left(1-\frac{x}{3}\right)}$ × $\frac{9}{\left(x+\frac{1}{3}\right)\left(1+\frac{x}{3}\right)}$ is _________.

Question

The simplified form of the expression $\frac{\left(x^2-9\right) &times;\left(9 x^2-1\right)}{\left(x-\frac{1}{3}\right) &times;\left(1-\frac{x}{3}\right)}$ &times; $\frac{9}{\left(x+\frac{1}{3}\right)\left(1+\frac{x}{3}\right)}$ is _________.

Prepp Team · Accepted Answer

Simplifying the Algebraic Expression

The problem asks us to simplify a complex algebraic expression involving products and fractions. We need to break down the expression into smaller parts and simplify each part before combining them.

The given expression is:

$\frac{\left(x^2-9\right) \times\left(9 x^2-1\right)}{\left(x-\frac{1}{3}\right) \times\left(1-\frac{x}{3}\right)}$ $\times$ $\frac{9}{\left(x+\frac{1}{3}\right)\left(1+\frac{x}{3}\right)}$

Step 1: Simplify the terms in the numerator of the first fraction

The numerator of the first fraction is $(x^2-9)(9x^2-1)$. Both terms are differences of squares:

$x^2 - 9 = x^2 - 3^2 = (x-3)(x+3)$
$9x^2 - 1 = (3x)^2 - 1^2 = (3x-1)(3x+1)$

So the numerator becomes $(x-3)(x+3)(3x-1)(3x+1)$.

Step 2: Simplify the terms in the denominator of the first fraction

The denominator of the first fraction is $\left(x-\frac{1}{3}\right) \times\left(1-\frac{x}{3}\right)$.

$x - \frac{1}{3} = \frac{3x}{3} - \frac{1}{3} = \frac{3x-1}{3}$
$1 - \frac{x}{3} = \frac{3}{3} - \frac{x}{3} = \frac{3-x}{3}$

The product is $\frac{3x-1}{3} \times \frac{3-x}{3} = \frac{(3x-1)(3-x)}{9}$. Note that $3-x = -(x-3)$, so this is $\frac{(3x-1)(-(x-3))}{9} = -\frac{(3x-1)(x-3)}{9}$.

Step 3: Write the simplified first fraction

The first fraction is $\frac{\text{Numerator}}{\text{Denominator}}$:

$\frac{(x-3)(x+3)(3x-1)(3x+1)}{-\frac{(3x-1)(x-3)}{9}}$

Dividing by a fraction is the same as multiplying by its reciprocal:

$(x-3)(x+3)(3x-1)(3x+1) \times \frac{-9}{(3x-1)(x-3)}$

Assuming $x \neq 3$ and $x \neq \frac{1}{3}$, we can cancel out the common terms $(x-3)$ and $(3x-1)$ from the numerator and denominator:

$(x+3)(3x+1) \times (-9)$

Step 4: Simplify the terms in the denominator of the second fraction

The denominator of the second fraction is $\left(x+\frac{1}{3}\right)\left(1+\frac{x}{3}\right)$.

$x + \frac{1}{3} = \frac{3x}{3} + \frac{1}{3} = \frac{3x+1}{3}$
$1 + \frac{x}{3} = \frac{3}{3} + \frac{x}{3} = \frac{3+x}{3}$

The product is $\frac{3x+1}{3} \times \frac{3+x}{3} = \frac{(3x+1)(3+x)}{9}$. Since $3+x = x+3$, this is $\frac{(3x+1)(x+3)}{9}$.

Step 5: Write the simplified second fraction

The second fraction is $\frac{9}{\text{Denominator}}$:

$\frac{9}{\frac{(3x+1)(x+3)}{9}}$

Multiply 9 by the reciprocal of the denominator:

$9 \times \frac{9}{(3x+1)(x+3)} = \frac{81}{(3x+1)(x+3)}$

Step 6: Multiply the simplified fractions

Now we multiply the simplified first fraction from Step 3 by the simplified second fraction from Step 5:

$\left((x+3)(3x+1) \times (-9)\right) \times \left(\frac{81}{(3x+1)(x+3)}\right)$

Rearrange the terms:

$(-9) \times 81 \times \frac{(x+3)(3x+1)}{(3x+1)(x+3)}$

Assuming $x \neq -3$ and $x \neq -\frac{1}{3}$, we can cancel out the common term $(x+3)(3x+1)$ from the numerator and denominator of the fraction part:

$(-9) \times 81 \times 1$

Step 7: Final Calculation

$-9 \times 81 = -729$

The simplified form of the expression is $-729$.

Revision Table: Key Simplification Steps

Part of Expression	Original Form	Simplified Form	Notes
Numerator 1st Fraction	$(x^2-9)(9x^2-1)$	$(x-3)(x+3)(3x-1)(3x+1)$	Difference of Squares
Denominator 1st Fraction	$\left(x-\frac{1}{3}\right)\left(1-\frac{x}{3}\right)$	$-\frac{(3x-1)(x-3)}{9}$	Common Denominators, Factor out -1
Denominator 2nd Fraction	$\left(x+\frac{1}{3}\right)\left(1+\frac{x}{3}\right)$	$\frac{(3x+1)(x+3)}{9}$	Common Denominators
1st Fraction Simplified	$\frac{(x^2-9)(9x^2-1)}{(x-\frac{1}{3})(1-\frac{x}{3})}$	$(x+3)(3x+1)(-9)$	Cancel terms
2nd Fraction Simplified	$\frac{9}{(x+\frac{1}{3})(1+\frac{x}{3})}$	$\frac{81}{(3x+1)(x+3)}$	Simplify fraction
Overall Simplified	Product of Simplified Fractions	$-729$	Cancel terms and multiply

Additional Information: Algebraic Identities and Rational Expressions

This problem extensively uses algebraic identities and properties of rational expressions. Understanding these concepts is crucial for simplifying such expressions.

Difference of Squares: An important identity is $a^2 - b^2 = (a-b)(a+b)$. This was used to factor $x^2-9$ and $9x^2-1$. Recognizing these patterns simplifies factorization.
Simplifying Rational Expressions: A rational expression is a fraction where the numerator and denominator are polynomials. Simplifying involves factoring the numerator and denominator and cancelling any common factors. This is only valid when the common factor is not zero.
Dividing by a Fraction: To divide by a fraction, you multiply by its reciprocal. The reciprocal of $\frac{a}{b}$ is $\frac{b}{a}$, assuming $a \neq 0$ and $b \neq 0$.
Combining Fractions: When adding or subtracting fractions, we find a common denominator. In this problem, converting terms like $x - \frac{1}{3}$ to $\frac{3x-1}{3}$ involves finding a common denominator (3).
Conditions for Simplification: When cancelling terms like $(x-3)$ or $(3x-1)$, we assume that these terms are not equal to zero. Therefore, the simplified expression $-729$ is valid for all $x$ except for the values that would make the original denominators zero ($x \neq \frac{1}{3}, x \neq 3, x \neq -\frac{1}{3}, x \neq -3$). However, the question asks for the simplified *form*, which is typically the result after cancelling terms, assuming the variables are not those specific values.

Mastering these techniques allows you to simplify complex algebraic expressions systematically.

Prepp Team · Answer

-81

Prepp Team · Answer

729

Prepp Team · Answer

81