Respuesta :
We are asked to simplify the following expression:
[tex]\sqrt[]{125}\cdot(\sqrt[]{5}-\sqrt[]{7})[/tex]We can begin with solving the square root of 125.
125 can be expressed as:
[tex]125=5\cdot5\cdot5[/tex]5 cubed is equal to 125.
Since we have a square root, it is convenient to express that 125 in squares:
[tex]125=5^2\cdot5[/tex]Then, for the square root of 125:
[tex]\sqrt[]{125}=\sqrt[]{5^2\cdot5}[/tex]The square root of a product is equivalent to the product of the square roots:
[tex]\sqrt[]{125}=\sqrt[]{5^2}\cdot\sqrt[]{5}[/tex]We know that the square root of 5 squared is 5:
[tex]\sqrt[]{5^2}=5[/tex]Then, we can say that:
[tex]\sqrt[]{125}=5\cdot\sqrt[]{5}[/tex]We can replace the term in the first equation:
[tex]\sqrt[]{125}\cdot(\sqrt[]{5}-\sqrt[]{7})=5\cdot\sqrt[]{5}\cdot(\sqrt[]{5}-\sqrt[]{7})[/tex]Now, we can solve the parenthesis:
[tex]5\cdot\sqrt[]{5}\cdot(\sqrt[]{5}-\sqrt[]{7})=5\cdot\sqrt[]{5}\cdot\sqrt[]{5}-5\cdot\sqrt[]{5}\cdot\sqrt[]{7}[/tex]The product of the square roots is equivalent to the square root of the product, then:
[tex]\begin{gathered} 5\cdot\sqrt[]{5\cdot5}-5\cdot\sqrt[]{5\cdot7} \\ 5\cdot\sqrt[]{25}-5\cdot\sqrt[]{35} \\ 5\cdot5-5\cdot\sqrt[]{35} \end{gathered}[/tex]Finally, the expression simplified will be:
[tex]25-5\cdot\sqrt[]{35}[/tex]