Question

1. $sqrt8{x^{16}y^{8}}$ 9. $sqrt4{81(x - 4)^{4}}$ 11. $sqrt4{a^{12}}$ 12. $sqrt3{a^{12}}$

Explanation:

Let's solve each problem one by one:

Problem 8: $\boldsymbol{\sqrt[8]{x^{16}y^{8}}}$

Step 1: Use the property of radicals $\sqrt[n]{a^m}=a^{\frac{m}{n}}$

For $x^{16}$: $\sqrt[8]{x^{16}} = x^{\frac{16}{8}}$
For $y^{8}$: $\sqrt[8]{y^{8}} = y^{\frac{8}{8}}$

Step 2: Simplify the exponents

$x^{\frac{16}{8}} = x^2$ and $y^{\frac{8}{8}} = y^1 = y$
So, $\sqrt[8]{x^{16}y^{8}} = x^2y$

Step 1: Simplify $\sqrt[4]{81}$ and $\sqrt[4]{(x - 4)^4}$ separately

$\sqrt[4]{81}=\sqrt[4]{3^4}=3$ (since $3^4 = 81$)
$\sqrt[4]{(x - 4)^4}=\vert x - 4\vert$ (because the fourth root of a fourth power gives the absolute value to ensure non - negativity)

Step 2: Combine the results

$\sqrt[4]{81(x - 4)^4}=3\vert x - 4\vert$

Step 1: Apply the radical - exponent property $\sqrt[n]{a^m}=a^{\frac{m}{n}}$

Here, $n = 4$ and $m = 12$, so $\sqrt[4]{a^{12}}=a^{\frac{12}{4}}$

Step 2: Simplify the exponent

$\frac{12}{4}=3$, so $a^{\frac{12}{4}}=a^3$

Answer:

$x^2y$

Problem 9: $\boldsymbol{\sqrt[4]{81(x - 4)^4}}$