Simplify: $\frac{x+3}{x²+3x+9}+ \frac{x-3}{x²-3x+9}-\frac{54}{x⁴+9x²+81}$

This is a class 10 Question From Simplification of Rational Expressions chapter of Unit Algebra (Mathematics). All the steps for the solutions are mentioned in the description below. If that's hard for you to navigate, you can always visit the facebook link given at the end of every posts. 

Solution:

Given,

= $\frac{x+3}{x²+3x+9}+ \frac{x-3}{x²-3x+9}-\frac{54}{x⁴+9x²+81}$

= $\frac{(x+3)(x²-3x+9)}{(x²+3x+9)(x²-3x+9)}+ \frac{(x-3)(x²+3x+9)}{(x²-3x+9)(x²+3x+9)}$
$-\frac{54}{x⁴+9x²+81}$

= $\frac{x³+3³}{(x²+9)²-(3x)²} +\frac{x³-3³}{(x²-9)²-(3x)²}- \frac{54}{x⁴+9x²+81)}$

= $\frac{x³+3³+x³-3³}{(x²+9)²-(3x)²} -\frac{54}{(x²)²+9x²+9²)}$

= $\frac{2x³}{(x²+9)²-(3x)²} -\frac{54}{(x²+9)²-18x²+9x²)}$

= $\frac{2x³}{(x²+9)²-(3x)²} -\frac{54}{(x²+9)²-(9x)²}$

= $\frac{2x³-54}{(x²+9)²-(3x)²}$

= $\frac{2(x³-27)}{(x²+3x+9)(x²-3x+9)}$

= $\frac{2(x-3)(x²+3x+9)}{(x²+3x+9)(x²-3x+9)}$

= $\frac{2(x-3)}{x²-3x+9}$


= Answer


Explanation to the above answer.


Step 1: Copy the same question given.

Step 2: If we want to take the LCM between 3 and 4. We multiply both. We do the same here for the first two terms.

Step 3: In the numerator, we have (a+b)(a²-ab+b²) which is equal to a³+b³. Also, (a-b)(a²+ab+b²) which is equal to a³-b³.

Step 4: We have similar denominators. So, add the two numerator. 

Step 5: In the denominator of the second term, we have a⁴+b⁴ which can also be written as (a²+b²)²-2a²b².

Step 6: Write the denominator in the form of a²-b² and match the denominator of the first term. 

Step 7: Subtract the two terms 

Step 8: Take the 2 as a common factor in the numerator. In the denominator, we have a²-b² which we can write as (a+b)(a-b).

Step 9: Expand the formula (a³-b³) as (a-b)(a²+ab+b²).

Step 10: The term (x²+3x+9) in the numerator and denominator get cancelled and the remaining expression is our answer.


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Question: Simplify: (x+3)/(x²+3x+9) + (x-3)/(x²-3x+9) - 54/(x⁴+9x²+81) | Simplification of Rational Expressions | SciPiPupil

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