Question

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Answer 1

Given:

The expression is:

$\dfrac{1}{a+b}+\dfrac{2a}{a^2+b^2}+\dfrac{4a^7+4a^3b^4}{b^8-a^8}$

To find:

The simplified form of the given expression

Solution:

Formula used:

$x^2-y^2=(x-y)(x+y)$

We have,

$\dfrac{1}{a+b}+\dfrac{2a}{a^2+b^2}+\dfrac{4a^7+4a^3b^4}{b^8-a^8}$

It can be written as:

$=\dfrac{1}{a+b}+\dfrac{2a}{a^2+b^2}+\dfrac{4a^3(a^4+b^4)}{(b^4)^2-(a^4)^2}$

$=\dfrac{1}{a+b}+\dfrac{2a}{a^2+b^2}+\dfrac{4a^3(a^4+b^4)}{(b^4-a^4)(b^4+a^4)}$

$=\dfrac{1}{a+b}+\dfrac{2a}{a^2+b^2}+\dfrac{4a^3}{(b^2-a^2)(b^2+a^2)}$

$=\dfrac{1}{a+b}+\dfrac{2a}{a^2+b^2}+\dfrac{4a^3}{(b-a)(b+a)(b^2+a^2)}$

Taking LCM, we get

$=\dfrac{1(b-a)(b^2+a^2)+2a(b-a)(b+a)+4a^3}{(b-a)(b+a)(b^2+a^2)}$

$=\dfrac{b^3+a^2b-ab^2-a^3+2a(b^2-a^2)+4a^3}{(b-a)(b+a)(b^2+a^2)}$

$=\dfrac{b^3+a^2b-ab^2-a^3+2ab^2-2a^3+4a^3}{(b-a)(b+a)(b^2+a^2)}$

$=\dfrac{b^3+a^2b+ab^2+a^3}{(b-a)(b+a)(b^2+a^2)}$

Using the grouping method factories the numerator.

$=\dfrac{b(b^2+a^2)+a(b^2+a^2)}{(b-a)(b+a)(b^2+a^2)}$

$=\dfrac{(b+a)(b^2+a^2)}{(b-a)(b+a)(b^2+a^2)}$

Cancel out the common factors.

$=\dfrac{1}{b-a}$

Therefore, the value of the given expression is $\dfrac{1}{b-a}$.