1.2 Simplify the following completely: \( 1.2 .1 \quad\left(4 a^{2}-2 a b+b^{2}\right)(2 a+b) \) \( 1.2 .2 \quad\left(8 x^{12}\right)^{-\frac{1}{3}} \times 2 x^{-4}-x^{-8} \) \( 1.2 .3 \quad \frac{7^{2 n+2}-3.7^{2 n+1}}{49^{n} .4} \)

To simplify these expressions completely, let's break them down one by one: **1.2.1** \(\left(4 a^{2}-2 a b+b^{2}\right)(2 a+b)\) Notice that \(4 a^{2}-2 a b+b^{2}\) can be factored as \((2 a-b)^{2}\). Therefore, we have: \((2 a-b)^{2}(2 a+b) = (2 a-b)(2 a-b)(2 a+b)\). Now we can apply the distributive property (also known as the FOIL method for binomials): \((2 a-b)((2 a-b)(2 a+b)) = (2 a-b)(4 a^{2}-b^{2}) = 8 a^{3} - 4 a^{2} b - 4 a b^{2} + b^{3}\), which gives the final result of \(8 a^{3} - 4 a^{2} b - 4 a b^{2} + b^{3}\). **1.2.2** \(\left(8 x^{12}\right)^{-\frac{1}{3}} \times 2 x^{-4}-x^{-8}\) First, simplify \(\left(8 x^{12}\right)^{-\frac{1}{3}} = \frac{1}{(8 x^{12})^{\frac{1}{3}}} = \frac{1}{2 x^{4}}\). So, the expression becomes: \(\frac{1}{2 x^{4}} \times 2 x^{-4} - x^{-8} = x^{-4} - x^{-8}\). Now, write \(x^{-4}\) as \(x^{-4} - x^{-8}\) which can be factored: \(x^{-4}(1 - x^{-4})\), resulting in \(\frac{x^{-4}(x^{4} - 1)}{x^{4}} = \frac{(x^{4} - 1)}{x^{8}}\). **1.2.3** \(\frac{7^{2 n+2}-3.7^{2 n+1}}{49^{n} .4}\) Rewriting \(49^{n}\) as \((7^{2})^{n} = 7^{2n}\), we have: \(\frac{7^{2n+2} - 3 \cdot 7^{2n+1}}{7^{2n} \cdot 4}\). Factoring \(7^{2n+1}\) out from the numerator yields: \(\frac{7^{2n+1}(7 - 3)}{4 \cdot 7^{2n}} = \frac{4 \cdot 7^{2n+1}}{4 \cdot 7^{2n}}\). Finally, simplifying gives us: \(\frac{4 \cdot 7^{2n+1}}{4 \cdot 7^{2n}} = 7^{1} = 7\). Thus, the final simplified forms are: 1. \(\quad 8 a^{3} - 4 a^{2} b - 4 a b^{2} + b^{3}\) 2. \(\quad \frac{(x^{4}-1)}{x^{8}}\) 3. \(\quad 7\)