Certainly! Let's evaluate each expression step-by-step using appropriate identities.
### Part (i): Evaluate [tex]\((a^2 + b^2)^2\)[/tex]
To evaluate [tex]\((a^2 + b^2)^2\)[/tex], we can use the identity for the square of a binomial:
[tex]\[
(p + q)^2 = p^2 + 2pq + q^2
\][/tex]
In this case, let [tex]\(p = a^2\)[/tex] and [tex]\(q = b^2\)[/tex]. Substituting these into the identity, we get:
[tex]\[
(a^2 + b^2)^2 = (a^2)^2 + 2(a^2)(b^2) + (b^2)^2
\][/tex]
Now, simplifying each term:
[tex]\[
(a^2)^2 = a^4
\][/tex]
[tex]\[
2(a^2)(b^2) = 2a^2b^2
\][/tex]
[tex]\[
(b^2)^2 = b^4
\][/tex]
Combining these results, we have:
[tex]\[
(a^2 + b^2)^2 = a^4 + 2a^2b^2 + b^4
\][/tex]
For specific values [tex]\(a = 2\)[/tex] and [tex]\(b = 3\)[/tex]:
[tex]\[
a^4 = 2^4 = 16
\][/tex]
[tex]\[
2a^2b^2 = 2 \cdot 2^2 \cdot 3^2 = 2 \cdot 4 \cdot 9 = 72
\][/tex]
[tex]\[
b^4 = 3^4 = 81
\][/tex]
Adding these results together:
[tex]\[
16 + 72 + 81 = 169
\][/tex]
So, the value of [tex]\((a^2 + b^2)^2\)[/tex] is [tex]\(\boxed{169}\)[/tex].
### Part (ii): Evaluate [tex]\((xy + 2z)^2\)[/tex]
To evaluate [tex]\((xy + 2z)^2\)[/tex], we can again use the identity for the square of a binomial:
[tex]\[
(p + q)^2 = p^2 + 2pq + q^2
\][/tex]
In this case, let [tex]\(p = xy\)[/tex] and [tex]\(q = 2z\)[/tex]. Substituting these into the identity, we get:
[tex]\[
(xy + 2z)^2 = (xy)^2 + 2(xy)(2z) + (2z)^2
\][/tex]
Now, simplifying each term:
[tex]\[
(xy)^2 = x^2 y^2
\][/tex]
[tex]\[
2(xy)(2z) = 4xyz
\][/tex]
[tex]\[
(2z)^2 = 4z^2
\][/tex]
Combining these results, we have:
[tex]\[
(xy + 2z)^2 = x^2 y^2 + 4xyz + 4z^2
\][/tex]
For specific values [tex]\(x = 1\)[/tex], [tex]\(y = 2\)[/tex], and [tex]\(z = 3\)[/tex]:
[tex]\[
x^2 y^2 = 1^2 \cdot 2^2 = 1 \cdot 4 = 4
\][/tex]
[tex]\[
4xyz = 4 \cdot 1 \cdot 2 \cdot 3 = 24
\][/tex]
[tex]\[
4z^2 = 4 \cdot 3^2 = 4 \cdot 9 = 36
\][/tex]
Adding these results together:
[tex]\[
4 + 24 + 36 = 64
\][/tex]
So, the value of [tex]\((xy + 2z)^2\)[/tex] is [tex]\(\boxed{64}\)[/tex].
