Simplify ( cube root of 81x)/( cube root of y)
The question asks for the simplification of a fraction where both the numerator and the denominator contain cube roots. Specifically, it wants you to simplify the expression that results from dividing the cube root of 81 times a variable x by the cube root of another variable y. This task likely involves applying the properties of radicals and possibly algebraic rules for simplifying expressions.
$\frac{\sqrt[3]{81 x}}{\sqrt[3]{y}}$
Refine the numerator.
Express $81x$ as $3^3 \cdot (3x)$.
Extract $27$ from $81$. $\frac{\sqrt[3]{27(3)x}}{\sqrt[3]{y}}$
Represent $27$ as $3^3$. $\frac{\sqrt[3]{3^3 \cdot 3x}}{\sqrt[3]{y}}$
Introduce parentheses. $\frac{\sqrt[3]{3^3 \cdot (3x)}}{\sqrt[3]{y}}$
Extract terms from the radical. $\frac{3\sqrt[3]{3x}}{\sqrt[3]{y}}$
Multiply $\frac{3\sqrt[3]{3x}}{\sqrt[3]{y}}$ by $\frac{(\sqrt[3]{y})^2}{(\sqrt[3]{y})^2}$.
Simplify the denominator.
Multiply $\frac{3\sqrt[3]{3x}}{\sqrt[3]{y}}$ by $\frac{(\sqrt[3]{y})^2}{(\sqrt[3]{y})^2}$.
Elevate $\sqrt[3]{y}$ to the power of $1$. $\frac{3\sqrt[3]{3x}(\sqrt[3]{y})^2}{(\sqrt[3]{y})^1(\sqrt[3]{y})^2}$
Apply the exponent combination rule $a^m a^n = a^{m+n}$. $\frac{3\sqrt[3]{3x}(\sqrt[3]{y})^2}{(\sqrt[3]{y})^{1+2}}$
Sum $1$ and $2$. $\frac{3\sqrt[3]{3x}(\sqrt[3]{y})^2}{(\sqrt[3]{y})^3}$
Transform $(\sqrt[3]{y})^3$ into $y$.
Rewrite $\sqrt[3]{y}$ using $y^{\frac{1}{3}}$. $\frac{3\sqrt[3]{3x}(\sqrt[3]{y})^2}{((y^{\frac{1}{3}}))^3}$
Utilize the power rule $(a^m)^n = a^{mn}$. $\frac{3\sqrt[3]{3x}(\sqrt[3]{y})^2}{y^{\frac{1}{3} \cdot 3}}$
Combine $\frac{1}{3}$ and $3$. $\frac{3\sqrt[3]{3x}(\sqrt[3]{y})^2}{y^{\frac{3}{3}}}$
Eliminate the common factor of $3$.
Remove the common factor. $\frac{3\sqrt[3]{3x}(\sqrt[3]{y})^2}{y^{\frac{\cancel{3}}{\cancel{3}}}}$
Rephrase the expression. $\frac{3\sqrt[3]{3x}(\sqrt[3]{y})^2}{y^1}$
Condense the expression. $\frac{3\sqrt[3]{3x}(\sqrt[3]{y})^2}{y}$
Refactor the numerator.
Reformulate $(\sqrt[3]{y})^2$ as $\sqrt[3]{y^2}$. $\frac{3\sqrt[3]{3x}\sqrt[3]{y^2}}{y}$
Merge using the radical product rule. $\frac{3\sqrt[3]{y^2 \cdot 3x}}{y}$
Reposition $3$ to precede $y^2$. $\frac{3\sqrt[3]{3y^2x}}{y}$
The problem involves simplifying a radical expression with cube roots. Here are the relevant knowledge points and detailed explanations:
Cube Roots: The cube root of a number $a$, denoted as $\sqrt[3]{a}$, is a number that, when raised to the third power, gives $a$. That is, if $b = \sqrt[3]{a}$, then $b^3 = a$.
Properties of Exponents: When simplifying expressions with exponents, several rules are used:
Product Rule: $a^m \cdot a^n = a^{m+n}$.
Power Rule: $(a^m)^n = a^{mn}$.
Radical and Exponent Relation: $\sqrt[n]{a^m} = a^{\frac{m}{n}}$.
Simplifying Radicals: When simplifying radicals, like terms under the radical can be combined, and factors that are perfect powers of the radical index can be taken out of the radical.
Rationalizing the Denominator: To avoid having a radical in the denominator, we can multiply the numerator and denominator by an appropriate form of 1 (like $(\sqrt[3]{y})^2/(\sqrt[3]{y})^2$) to eliminate the radical in the denominator.
Combining Radicals: Radicals with the same index and radicand (the number under the radical symbol) can be combined using the product rule for radicals.
By applying these principles, we can simplify the given expression to a more elementary form without radicals in the denominator.