9) $sqrt{8x^7y}$ 10) $sqrt{54x^3y^8}$ 11) $sqrt{90x^{10}y^{20}}$ 12) $sqrt{172x^5y^{13}}$ 13) $sqrt3{8x^7y}$ 14) $sqrt3{54x^3y^8}$ 15) $sqrt3{90x^{10}y^{20}}$ 16) $sqrt3{135x^5y^{13}}$

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Accepted Answer

Let's solve these radical simplification problems one by one. We'll start with the square roots (problems 9 - 12) and then the cube roots (problems 13 - 16).

### Problem 9: Simplify $\sqrt{8x^7y}$
# Explanation:
## Step 1: Factor the radicand into perfect squares and remaining factors
We know that $8 = 4\times2$, $x^7 = x^6\times x$, and $y$ is just $y$. So we can rewrite the radicand as:
$8x^7y = 4\times2\times x^6\times x\times y$
## Step 2: Take the square root of the perfect squares
$\sqrt{4\times2\times x^6\times x\times y}=\sqrt{4}\times\sqrt{x^6}\times\sqrt{2xy}$
Since $\sqrt{4} = 2$ and $\sqrt{x^6}=x^3$ (because $(x^3)^2 = x^6$), we get:
$2\times x^3\times\sqrt{2xy}=2x^3\sqrt{2xy}$
# Answer:
$2x^3\sqrt{2xy}$

### Problem 10: Simplify $\sqrt{54x^3y^8}$
# Explanation:
## Step 1: Factor the radicand into perfect squares and remaining factors
$54 = 9\times6$, $x^3 = x^2\times x$, and $y^8=(y^4)^2$. So:
$54x^3y^8 = 9\times6\times x^2\times x\times(y^4)^2$
## Step 2: Take the square root of the perfect squares
$\sqrt{9\times6\times x^2\times x\times(y^4)^2}=\sqrt{9}\times\sqrt{x^2}\times\sqrt{(y^4)^2}\times\sqrt{6x}$
We know that $\sqrt{9} = 3$, $\sqrt{x^2}=x$, and $\sqrt{(y^4)^2}=y^4$. So:
$3\times x\times y^4\times\sqrt{6x}=3xy^4\sqrt{6x}$
# Answer:
$3xy^4\sqrt{6x}$

### Problem 11: Simplify $\sqrt{90x^{10}y^{20}}$
# Explanation:
## Step 1: Factor the radicand into perfect squares and remaining factors
$90 = 9\times10$, $x^{10}=(x^5)^2$, and $y^{20}=(y^{10})^2$. So:
$90x^{10}y^{20}=9\times10\times(x^5)^2\times(y^{10})^2$
## Step 2: Take the square root of the perfect squares
$\sqrt{9\times10\times(x^5)^2\times(y^{10})^2}=\sqrt{9}\times\sqrt{(x^5)^2}\times\sqrt{(y^{10})^2}\times\sqrt{10}$
Since $\sqrt{9} = 3$, $\sqrt{(x^5)^2}=x^5$, and $\sqrt{(y^{10})^2}=y^{10}$, we get:
$3\times x^5\times y^{10}\times\sqrt{10}=3x^5y^{10}\sqrt{10}$
# Answer:
$3x^5y^{10}\sqrt{10}$

### Problem 12: Simplify $\sqrt{172x^5y^{13}}$
# Explanation:
## Step 1: Factor the radicand into perfect squares and remaining factors
$172 = 4\times43$, $x^5 = x^4\times x$, and $y^{13}=y^{12}\times y=(y^6)^2\times y$. So:
$172x^5y^{13}=4\times43\times x^4\times x\times y^{12}\times y$
## Step 2: Take the square root of the perfect squares
$\sqrt{4\times43\times x^4\times x\times y^{12}\times y}=\sqrt{4}\times\sqrt{x^4}\times\sqrt{y^{12}}\times\sqrt{43xy}$
We know that $\sqrt{4} = 2$, $\sqrt{x^4}=x^2$, and $\sqrt{y^{12}}=y^6$. So:
$2\times x^2\times y^6\times\sqrt{43xy}=2x^2y^6\sqrt{43xy}$
# Answer:
$2x^2y^6\sqrt{43xy}$

### Problem 13: Simplify $\sqrt[3]{8x^7y}$
# Explanation:
## Step 1: Factor the radicand into perfect cubes and remaining factors
$8 = 2^3$, $x^7 = x^6\times x=(x^2)^3\times x$, and $y$ is just $y$. So:
$8x^7y = 2^3\times(x^2)^3\times x\times y$
## Step 2: Take the cube root of the perfect cubes
$\sqrt[3]{2^3\times(x^2)^3\times x\times y}=\sqrt[3]{2^3}\times\sqrt[3]{(x^2)^3}\times\sqrt[3]{xy}$
Since $\sqrt[3]{2^3}=2$ and $\sqrt[3]{(x^2)^3}=x^2$, we get:
$2\times x^2\times\sqrt[3]{xy}=2x^2\sqrt[3]{xy}$
# Answer:
$2x^2\sqrt[3]{xy}$

### Problem 14: Simplify $\sqrt[3]{54x^3y^8}$
# Explanation:
## Step 1: Factor the radicand into perfect cubes and remaining factors
$54 = 27\times2 = 3^3\times2$, $x^3$ is a perfect cube, and $y^8 = y^6\times y^2=(y^2)^3\times y^2$. So:
$54x^3y^8 = 3^3\times2\times x^3\times(y^2)^3\times y^2$
## Step 2: Take the cube root of the perfect cubes
$\sqrt[3]{3^3\times2\times x^3\times(y^2)^3\times y^2}=\sqrt[3]{3^3}\times\sqrt[3]{x^3}\times\sqrt[3]{(y^2)^3}\times\sqrt[3]{2y^2}$
Since $\sqrt[3]{3^3}=3$, $\sqrt[3]{x^3}=x$, and $\sqrt[3]{(y^2)^3}=y^2$, we get:
$3\times x\times y^2\times\sqrt[3]{2y^2}=3xy^2\sqrt[3]{2y^2}$
# Answer:
$3xy^2\sqrt[3]{2y^2}$

### Problem 15: Simplify $\sqrt[3]{90x^{10}y^{20}}$
# Explanation:
## Step 1: Factor the radicand into perfect cubes and remaining factors
$90 = 27\times\frac{10}{3}$? No, better to factor as $90 = 27\times\frac{10}{3}$ is not helpful. Wait, $90 = 27\times \frac{10}{3}$ is wrong. Let's do prime factorization: $90=2\times3^3\times5$. $x^{10}=x^9\times x=(x^3)^3\times x$, $y^{20}=y^{18}\times y^2=(y^6)^3\times y^2$. So:
$90x^{10}y^{20}=2\times3^3\times5\times(x^3)^3\times x\times(y^6)^3\times y^2$
## Step 2: Take the cube root of the perfect cubes
$\sqrt[3]{2\times3^3\times5\times(x^3)^3\times x\times(y^6)^3\times y^2}=\sqrt[3]{3^3}\times\sqrt[3]{(x^3)^3}\times\sqrt[3]{(y^6)^3}\times\sqrt[3]{2\times5\times x\times y^2}$
Simplify the perfect cube roots: $\sqrt[3]{3^3}=3$, $\sqrt[3]{(x^3)^3}=x^3$, $\sqrt[3]{(y^6)^3}=y^6$. So:
$3\times x^3\times y^6\times\sqrt[3]{10xy^2}=3x^3y^6\sqrt[3]{10xy^2}$
# Answer:
$3x^3y^6\sqrt[3]{10xy^2}$

### Problem 16: Simplify $\sqrt[3]{135x^5y^{13}}$
# Explanation:
## Step 1: Factor the radicand into perfect cubes and remaining factors
$135 = 27\times5 = 3^3\times5$, $x^5 = x^3\times x^2$, $y^{13}=y^{12}\times y=(y^4)^3\times y$. So:
$135x^5y^{13}=3^3\times5\times x^3\times x^2\times(y^4)^3\times y$
## Step 2: Take the cube root of the perfect cubes
$\sqrt[3]{3^3\times5\times x^3\times x^2\times(y^4)^3\times y}=\sqrt[3]{3^3}\times\sqrt[3]{x^3}\times\sqrt[3]{(y^4)^3}\times\sqrt[3]{5x^2y}$
Simplify the perfect cube roots: $\sqrt[3]{3^3}=3$, $\sqrt[3]{x^3}=x$, $\sqrt[3]{(y^4)^3}=y^4$. So:
$3\times x\times y^4\times\sqrt[3]{5x^2y}=3xy^4\sqrt[3]{5x^2y}$
# Answer:
$3xy^4\sqrt[3]{5x^2y}$

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QUESTION IMAGE

Question

Explanation:

Problem 9: Simplify \(\sqrt{8x^7y}\)

Step 1: Factor the radicand into perfect squares and remaining factors

Step 2: Take the square root of the perfect squares

Step 1: Factor the radicand into perfect squares and remaining factors

Step 2: Take the square root of the perfect squares

Step 1: Factor the radicand into perfect squares and remaining factors

Step 2: Take the square root of the perfect squares

Answer:

Problem 10: Simplify \(\sqrt{54x^3y^8}\)