Please provide a solution on how the radical expression turned into that simplified form in that image. Ty

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Adithya027 Adithya027 · Answer 1 · 2024-01-13T09:57:44+01:00

Step-by-step explanation:

[tex]\sqrt[3]{7}[/tex]÷[tex]\sqrt[3]{3}[/tex]

[tex]7^{\frac{1}{3} }[/tex]÷[tex]3^{\frac{1}{3} }[/tex]

[tex](\frac{7}{3})^{\frac{1}{3}}[/tex]

[tex](\frac{7}{3})^{\frac{1}{3}}[/tex]×1

[tex](\frac{7}{3})^{\frac{1}{3}}[/tex]×[tex](\frac{9}{9})^\frac{1}{3}[/tex]

[tex](\frac{7X9}{3X9})^{\frac{1}{3}}[/tex]

[tex](\frac{63}{27})^{\frac{1}{3}}[/tex]

[tex]\sqrt[3]{63}/3[/tex] [[tex]\sqrt[3]{27}=3[/tex]]

∴ [tex]\sqrt[3]{7}[/tex]÷[tex]\sqrt[3]{3}[/tex]=[tex]\sqrt[3]{63}/3[/tex]

jimrgrant1 jimrgrant1 · Answer 2 · 2024-01-13T10:00:02+01:00

Answer:

see explanation

Step-by-step explanation:

Using the property of radicals

• [tex]\sqrt[n]{a}[/tex] × [tex]\sqrt[n]{b}[/tex] = [tex]\sqrt[n]{ab}[/tex]

given

[tex]\sqrt[3]{7}[/tex] ÷ [tex]\sqrt[3]{3}[/tex]

= [tex]\frac{\sqrt[3]{7} }{\sqrt[3]{3} }[/tex]

multiply numerator and denominator by [tex]\sqrt[3]{9}[/tex]

= [tex]\frac{\sqrt[3]{7}(\sqrt[3]{9}) }{\sqrt[3]{3}(\sqrt[3]{9}) }[/tex]

= [tex]\frac{\sqrt[3]{63} }{\sqrt[3]{27} }[/tex]

= [tex]\frac{\sqrt[3]{63} }{3}[/tex]