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Sagot :
For this exercise, we must use the binomial theorem. The formula of this theorem is described as follows
[tex](a+b)^n=\sum_{i\mathop{=}0}^nnCi(a^{n-i})(b^i)[/tex]Where nCi represents the combinatory of n between i. For our case,
[tex]a=a^2\text{ y }b=b[/tex]Replacing the values you have in the formula
[tex]\begin{gathered} (a^2+b)^8=\sum_{i\mathop{=}0}^88Ci(a^2)^{8-i}(b^i) \\ =\frac{8!}{0!(8-0)!}(a^2)^{8-0}b^0+\frac{8!}{1!(8-1)!}(a^2)^{8-1}b^1+\frac{8!}{2!(8-2)!}(a^2)^{8-2}b^2+....+\frac{8!}{8!(8-8)!}(a^2)^{8-8}b^8 \\ =\frac{8!}{8!}(a^2)^8+\frac{8!}{1(7)!}(a^2)^7b^1+\frac{8!}{2(6)!}(a^2)^6b^2+....+\frac{8!}{8!}(a^2)^5b^8 \\ =a^{16}+8a^{14}b+28a^{12}b^2+56a^{10}b^3+70a^8b^4+56a^6b^5+28a^4b^6+8a^2b^7+b^8 \end{gathered}[/tex]Thus,the expansion and simplification is as follows
[tex](a^2+b)^8=\placeholder{⬚}+8a^{14}b+28a^{12}b^2+56a^{10}b^3+70a^8b^4+56a^6b^5+28a^4b^6+8a^2b^7+b^8[/tex]
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