Simplify: $\frac{3x-1}{9x²-3x+1}- \frac{3x+1}{9x²+3x+1}+\frac{54x³}{81x⁴+9x²+1}$

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,

= $\frac{3x-1}{9x²-3x+1}- \frac{3x+1}{9x²+3x+1}+\frac{54x³}{81x⁴+9x²+1}$

= $\frac{(3x-1)(9x²+3x+1)}{(9x²-3x+1)(9x²+3x+1)}- \frac{(3x+1)(9x²-3x+1)}{(9x²-3x+1)(9x²+3x+1)}$
$+\frac{54x³}{81x⁴+9x²+1}$

= $\frac{(3x-1)(9x²+3x+1) - (3x+1)(9x²-3x+1)}{(9x²-3x+1)(9x²+3x+1)}+\frac{54x³}{81x⁴+9x²+1}$

= $\frac{27x³-1-27x³-1}{(9x²+1)²-(3x)²}+\frac{54x³}{(9x²+1)²+9x²}$

= $\frac{-2}{(9x²+1)²-(3x)²}+\frac{54x³}{(9x²-1)²-(3x)²}$

= $\frac{54x³-2}{(9x²+1)²-(3x)²}$

= $\frac{2(3x-1)(9x²+3x+1)}{(9x²+3x+1)(9x²-3x+1)}$

= $\frac{2(3x-1}{9x²-3x+1}$

= 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: Since we have common denominators, we subtract the two terms. 

Step 4: Now, we multiply the factors and write their result in expressions. In the numerator, we have (a-b)(a²+ab+b²) = (a³-b³) and (a+b)(a²-ab+b²) = (a³+b³). In the denominator, we have 81x⁴+9x²+1 = (9x²)²+9x²+(1)² = (9x²+1)²-18x²+9x² = (9x²+1)²-(3x)².

Step 5: Perform the required operations in the numerator as well as denominator.

Step 6: Combine the two terms because the Denominators are equal. 

Step 7: Take 2 as the common factor and expand the terms (a³-b³) as (a-b)(a²+ab+b²) in the numerator. 

Step 8: Write your answer by eliminating common terms in the numerator and denominator. 



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Question: Simplify: (3x-1)/(9x²-3x+) -(3x+1)/(9x²+3x+1) +(54x³)/(81x⁴+9x²+1) | Simplification of Rational Expressions | SciPiPupil

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