Question

a5+b5/a+b = , without actual division algebra​

Answer 1

Answer:

$a^{4}-a^{3}b+a^{2}b^{2}-ab^{3}+b^{4}$

Step-by-step explanation:

We have to find the value of $\frac{(a^{5}+b^{5})}{a+b}$, without actual division algebra.

Clearly putting a=-b the expression $(a^{5}+b^{5})$ becomes zero.

So, (a+b) is definitely a factor of $(a^{5}+b^{5})$ term.

Now we can express $(a^{5}+b^{5})$ in the following form:

$(a^{5}+b^{5})$

= $(a^{5}+a^{4}b)-(a^{4}b+a^{3}b^{2})+(a^{3}b^{2}+a^{2}b^{3})-(a^{2}b^{3}+ab^{4})+(ab^{4}+b^{5})$

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

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

So, it can be written that,

$\frac{(a^{5}+b^{5})}{a+b}=a^{4}-a^{3}b+a^{2}b^{2}-ab^{3}+b^{4}$

(Answer)