Pierwiastki.
[tex]\huge\boxed{\sqrt[3]{27}-\sqrt[3]{64}+\sqrt[3]8=1}\\\boxed{\sqrt[3]{125}-\sqrt[3]{27}=2}\\\boxed{\left(4\sqrt[3]2\right)^3=128}\\\boxed{\left(3\sqrt[3]{125}\right)^3=3375}\\\boxed{\left(\dfrac{\sqrt[3]5}{10}\right)^3=\dfrac{1}{200}}[/tex]
ROZWIĄZANIA:
Definicja pierwiastka sześciennego:
[tex]\sqrt[3]a=b\iff b^3=a[/tex]
Twierdzenia:
[tex]\left(\sqrt[3]a\right)^3=a\\\\(a\cdot b)^n=a^n\cdot b^n\\\\\left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\qquad\text{dla}\ b\neq0[/tex]
[tex]\sqrt[3]{27}=3\ \text{bo}\ 3^3=27\\\sqrt[3]{64}=4\ \text{bo}\ 4^3=64\\\sqrt[3]8=2\ \text{bo}\ 2^3=8[/tex]
stąd mamy:
[tex]\sqrt[3]{27}-\sqrt[3]{64}+\sqrt[3]8=3-4+2=\boxed{1}[/tex]
[tex]\sqrt[3]{125}=5\ \text{bo}\ 5^3=125\\\sqrt[3]{27}=3\ \text{bo}\ 3^3=27[/tex]
stąd mamy:
[tex]\sqrt[3]{125}-\sqrt[3]{27}=5-3=\boxed{2}[/tex]
[tex]\left(4\sqrt[3]2\right)^3=4^3\cdot\left(\sqrt[3]2\right)^3=64\cdot2=\boxed{128}\\\\\left(3\sqrt[3]{125}\right)^3=3^3\cdot\left(\sqrt[3]{125}\right)^3=27\cdot125=\boxed{3375}\\\\\left(\dfrac{\sqrt[3]5}{10}\right)^3=\dfrac{\left(\sqrt[3]5\right)^3}{10^3}=\dfrac{5}{1000}=\boxed{\dfrac{1}{200}}[/tex]
