Sure, let's evaluate each of the given expressions using appropriate algebraic identities.
### (i) [tex]\(\left(a^2 + b^2\right)^2\)[/tex]
We can expand this using the identity for the square of a binomial [tex]\((u+v)^2 = u^2 + 2uv + v^2\)[/tex]. Here, [tex]\(u = a^2\)[/tex] and [tex]\(v = b^2\)[/tex]. Let's apply the identity:
[tex]\[
(a^2 + b^2)^2 = (a^2)^2 + 2(a^2)(b^2) + (b^2)^2
\][/tex]
Breaking it down step by step:
1. [tex]\((a^2)^2 = a^4\)[/tex]
2. [tex]\(2(a^2)(b^2) = 2a^2b^2\)[/tex]
3. [tex]\((b^2)^2 = b^4\)[/tex]
Putting it all together:
[tex]\[
(a^2 + b^2)^2 = a^4 + 2a^2b^2 + b^4
\][/tex]
### (ii) [tex]\((xy + 2z)^2\)[/tex]
Similarly, we can use the identity for the square of a binomial [tex]\((u+v)^2 = u^2 + 2uv + v^2\)[/tex]. Here, [tex]\(u = xy\)[/tex] and [tex]\(v = 2z\)[/tex]. Let's apply the identity:
[tex]\[
(xy + 2z)^2 = (xy)^2 + 2(xy)(2z) + (2z)^2
\][/tex]
Breaking it down step by step:
1. [tex]\((xy)^2 = x^2y^2\)[/tex]
2. [tex]\(2(xy)(2z) = 4xyz\)[/tex]
3. [tex]\((2z)^2 = 4z^2\)[/tex]
Putting it all together:
[tex]\[
(xy + 2z)^2 = x^2y^2 + 4xyz + 4z^2
\][/tex]
### (iii)
Unfortunately, for part (iii), no expression was provided, so we cannot solve it.
In summary:
1. [tex]\((a^2 + b^2)^2 = a^4 + 2a^2b^2 + b^4\)[/tex]
2. [tex]\((xy + 2z)^2 = x^2y^2 + 4xyz + 4z^2\)[/tex]
3. (Expression not provided for (iii) thus cannot be solved)
