Question

Expand and simplify (a^2+b)^8

Answer 1

For this exercise, we must use the binomial theorem. The formula of this theorem is described as follows

`(a+b)^n=\sum_{i\mathop{=}0}^nnCi(a^(n-i))(b^i)`

Where nCi represents the combinatory of n between i. For our case,

`a=a^2\text{ y }b=b`

Replacing the values you have in the formula

`\begin{gathered} (a^2+b)^8=\sum_{i\mathop{=}0}^88Ci(a^2)^(8-i)(b^i) \\ =(8!)/(0!(8-0)!)(a^2)^(8-0)b^0+(8!)/(1!(8-1)!)(a^2)^(8-1)b^1+(8!)/(2!(8-2)!)(a^2)^(8-2)b^2+....+(8!)/(8!(8-8)!)(a^2)^(8-8)b^8 \\ =(8!)/(8!)(a^2)^8+(8!)/(1(7)!)(a^2)^7b^1+(8!)/(2(6)!)(a^2)^6b^2+....+(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}`

Thus,the expansion and simplification is as follows

`(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`