4. \( -\sqrt[3]{-70 a^{18} b^{6}+6 a^{18} b^{6}} \) 5. \( \sqrt[3]{27 x^{12}}-\sqrt{25 x^{8}} \) 6. \( -\sqrt{100 x^{70} y^{26}} \) 7. \( \sqrt{81 x^{14} y^{2}} \) 8. \( -\sqrt[3]{-512 a^{27} b^{15}} \) 9. \( \sqrt{p^{100} q^{16}} \)

4. Factor out \(a^{18} b^{6}\): \[ -\sqrt[3]{-70a^{18} b^{6}+6a^{18} b^{6} } = -\sqrt[3]{(-70+6)a^{18}b^{6}} = -\sqrt[3]{-64a^{18}b^{6}}. \] Taking the cube root: \[ \sqrt[3]{-64} = -4,\quad \sqrt[3]{a^{18}} = a^6,\quad \sqrt[3]{b^{6}} = b^2. \] Thus, \[ -\sqrt[3]{-64a^{18}b^{6}} = -(-4a^6b^2) = 4a^6b^2. \] 5. Simplify each radical separately: \[ \sqrt[3]{27x^{12}} = \sqrt[3]{27}\cdot \sqrt[3]{x^{12}} = 3x^4, \] since \(27 = 3^3\) and \(x^{12} = \left(x^4\right)^3\). Also, \[ \sqrt{25x^{8}} = \sqrt{25}\cdot \sqrt{x^{8}} = 5x^4, \] because \(25 = 5^2\) and \(x^{8} = \left(x^4\right)^2\). Therefore, \[ \sqrt[3]{27x^{12}}-\sqrt{25x^{8}} = 3x^4-5x^4 = -2x^4. \] 6. Break up the square root: \[ -\sqrt{100x^{70}y^{26}} = -\sqrt{100}\cdot\sqrt{x^{70}}\cdot\sqrt{y^{26}}. \] Since \[ \sqrt{100} = 10,\quad x^{70}=(x^{35})^2,\quad y^{26}=(y^{13})^2, \] we have \[ \sqrt{x^{70}} = x^{35} \quad\text{and}\quad \sqrt{y^{26}} = y^{13}. \] Thus, \[ -\sqrt{100x^{70}y^{26}} = -10x^{35}y^{13}. \] 7. Factor the radicand: \[ \sqrt{81x^{14}y^{2}} = \sqrt{81}\cdot\sqrt{x^{14}}\cdot\sqrt{y^{2}}. \] Note that: \[ \sqrt{81} = 9,\quad x^{14}=(x^{7})^2,\quad y^{2}=(y)^2. \] So, \[ \sqrt{x^{14}} = x^7,\quad \sqrt{y^{2}} = y. \] Therefore, \[ \sqrt{81x^{14}y^{2}} = 9x^7y. \] 8. Evaluate the cube root: \[ -\sqrt[3]{-512a^{27}b^{15}}. \] Since: \[ \sqrt[3]{-512} = -8,\quad \sqrt[3]{a^{27}} = a^9,\quad \sqrt[3]{b^{15}} = b^5, \] the cube root becomes: \[ \sqrt[3]{-512a^{27}b^{15}} = -8a^9b^5. \] Including the negative sign outside, \[ -\sqrt[3]{-512a^{27}b^{15}} = -(-8a^9b^5) = 8a^9b^5. \] 9. Simplify the square root: \[ \sqrt{p^{100}q^{16}} = \sqrt{p^{100}}\cdot\sqrt{q^{16}}. \] Since: \[ \sqrt{p^{100}} = p^{50}\quad\text{and}\quad \sqrt{q^{16}} = q^8, \] we obtain: \[ \sqrt{p^{100}q^{16}} = p^{50}q^8. \]