Question

⁴√(256v⁷w¹²)

Explanation:

Step1: Rewrite the radical as a fractional exponent

The fourth root of a number can be written as raising that number to the power of $\frac{1}{4}$. So, $\sqrt[4]{256v^{7}w^{12}}=(256v^{7}w^{12})^{\frac{1}{4}}$.

Step2: Apply the power - of - a - product rule

The power - of - a - product rule states that $(ab)^n = a^n\times b^n$. So, we can rewrite the expression as $256^{\frac{1}{4}}\times(v^{7})^{\frac{1}{4}}\times(w^{12})^{\frac{1}{4}}$.

Step3: Simplify $256^{\frac{1}{4}}$

We know that $4^4 = 256$, so $256^{\frac{1}{4}}=\sqrt[4]{256} = 4$.

Step4: Simplify $(v^{7})^{\frac{1}{4}}$

Using the power - of - a - power rule $(a^m)^n=a^{m\times n}$, we have $(v^{7})^{\frac{1}{4}}=v^{\frac{7}{4}}=v^{1+\frac{3}{4}}=v\times v^{\frac{3}{4}}$.

Step5: Simplify $(w^{12})^{\frac{1}{4}}$

Using the power - of - a - power rule, $(w^{12})^{\frac{1}{4}}=w^{\frac{12}{4}} = w^{3}$.

Step6: Rewrite $v^{\frac{7}{4}}$ as a radical

$v^{\frac{7}{4}}=\sqrt[4]{v^{7}}=\sqrt[4]{v^{4}\times v^{3}}=v\sqrt[4]{v^{3}}$ (by the property $\sqrt[n]{ab}=\sqrt[n]{a}\times\sqrt[n]{b}$ for $a = v^{4}$ and $b = v^{3}$). But we can also write $v^{\frac{7}{4}}$ as $v^{1+\frac{3}{4}}=v\times v^{\frac{3}{4}}$, and $v^{\frac{3}{4}}=\sqrt[4]{v^{3}}$. However, we can also express the original expression in terms of exponents and simplify:

We know that $256 = 4^4$, so $256^{\frac{1}{4}}=4$.

For the variable $v$: $(v^{7})^{\frac{1}{4}}=v^{\frac{7}{4}}=v^{1+\frac{3}{4}}=v\times v^{\frac{3}{4}}$, and $v^{\frac{3}{4}}=\sqrt[4]{v^{3}}$, but we can also write the entire expression as:

$\sqrt[4]{256v^{7}w^{12}}=\sqrt[4]{256}\times\sqrt[4]{v^{7}}\times\sqrt[4]{w^{12}}=4\times\sqrt[4]{v^{4}\times v^{3}}\times w^{3}=4\times v\times\sqrt[4]{v^{3}}\times w^{3}=4vw^{3}\sqrt[4]{v^{3}}$

Or, if we want to express it with fractional exponents:

$\sqrt[4]{256v^{7}w^{12}} = 4v^{\frac{7}{4}}w^{3}=4v^{1+\frac{3}{4}}w^{3}=4vw^{3}v^{\frac{3}{4}}=4vw^{3}\sqrt[4]{v^{3}}$

Answer:

$4vw^{3}\sqrt[4]{v^{3}}$ (or $4v^{\frac{7}{4}}w^{3}$)