Exercise 7 Set 2 $$ \begin{array}{lll} ext { (1) } \sqrt[3]{64 x^{6} y^{12}} & ext { (4) }\left(\frac{2 x p}{6 x^{2}} ight)^{3} & ext { (7) } 6 x^{2} y^{4}+\left(-x y^{2} ight)^{2} \ ext { (2) } x^{-3} y & ext { (5) }-2\left(2 x y^{4} ight)^{2} & ext { (8) }\left(2^{-2} ight)^{2 x+1} \ ext { (3) }\left(\frac{1}{x} ight)^{-2} & ext { (6) } \frac{n}{-2 m^{-3}} & ext { (9) } \frac{-\sqrt[3]{27 x^{6}}}{\sqrt{9 x^{4} y^{10}}}\end{array} $$

Question

Exercise 7 Set 2 $$ \begin{array}{lll}	ext { (1) } \sqrt[3]{64 x^{6} y^{12}} & 	ext { (4) }\left(\frac{2 x p}{6 x^{2}}ight)^{3} & 	ext { (7) } 6 x^{2} y^{4}+\left(-x y^{2}ight)^{2} \ 	ext { (2) } x^{-3} y & 	ext { (5) }-2\left(2 x y^{4}ight)^{2} & 	ext { (8) }\left(2^{-2}ight)^{2 x+1} \ 	ext { (3) }\left(\frac{1}{x}ight)^{-2} & 	ext { (6) } \frac{n}{-2 m^{-3}} & 	ext { (9) } \frac{-\sqrt[3]{27 x^{6}}}{\sqrt{9 x^{4} y^{10}}}\end{array} $$

UpStudy AI · Accepted Answer

**(1)**  
We start with  
$$
\sqrt[3]{64x^6y^{12}}.
$$  
Notice that  
$$
64=4^3,\quad x^6=(x^2)^3,\quad y^{12}=(y^4)^3.
$$  
Thus, the cube root simplifies as  
$$
\sqrt[3]{4^3(x^2)^3(y^4)^3}=4x^2y^4.
$$

---

**(2)**  
The expression is  
$$
x^{-3}y.
$$  
Recall that $$x^{-3}=\frac{1}{x^3}$$. Therefore,  
$$
x^{-3}y=\frac{y}{x^3}.
$$

---

**(3)**  
We have  
$$
\left(\frac{1}{x}\right)^{-2}.
$$  
Raising to the $$-2$$ power gives  
$$
\left(\frac{1}{x}\right)^{-2}=x^2.
$$

---

**(4)**  
The expression is  
$$
\left(\frac{2xp}{6x^2}\right)^3.
$$  
First, simplify the fraction inside:  
$$
\frac{2xp}{6x^2}=\frac{2p}{6x}=\frac{p}{3x}.
$$  
Then, raising to the cube we get  
$$
\left(\frac{p}{3x}\right)^3=\frac{p^3}{27x^3}.
$$

---

**(5)**  
Given  
$$
-2\left(2xy^4\right)^2,
$$  
first compute the square:  
$$
\left(2xy^4\right)^2=4x^2y^8.
$$  
Then multiply by $$-2$$:  
$$
-2\cdot 4x^2y^8=-8x^2y^8.
$$

---

**(6)**  
We need to simplify  
$$
\frac{n}{-2m^{-3}}.
$$  
Recall that  
$$
m^{-3}=\frac{1}{m^3},
$$  
so the denominator becomes  
$$
-2m^{-3}=-\frac{2}{m^3}.
$$  
Thus,  
$$
\frac{n}{-\frac{2}{m^3}}=n\cdot\left(-\frac{m^3}{2}\right)=-\frac{nm^3}{2}.
$$

---

**(7)**  
Consider  
$$
6x^2y^4+(-xy^2)^2.
$$  
Square the second term:  
$$
(-xy^2)^2=(-1)^2\cdot (xy^2)^2=x^2y^4.
$$  
Now, adding the terms yields  
$$
6x^2y^4+x^2y^4=7x^2y^4.
$$

---

**(8)**  
We have  
$$
\left(2^{-2}\right)^{2x+1}.
$$  
Apply the power of a power rule:  
$$
\left(2^{-2}\right)^{2x+1}=2^{-2(2x+1)}=2^{-4x-2}.
$$

---

**(9)**  
The expression is  
$$
\frac{-\sqrt[3]{27x^6}}{\sqrt{9x^4y^{10}}}.
$$  
**Numerator:**  
$$
\sqrt[3]{27x^6}=\sqrt[3]{27}\cdot\sqrt[3]{x^6}=3x^2,
$$  
so the numerator becomes $$-3x^2$$.

**Denominator:**  
$$
\sqrt{9x^4y^{10}}=\sqrt{9}\cdot\sqrt{x^4}\cdot\sqrt{y^{10}}=3x^2\cdot y^5,
$$  
assuming all variables are positive.

Thus,  
$$
\frac{-3x^2}{3x^2y^5}=-\frac{1}{y^5}.
$$
**(1)** $$ 4x^2y^4 $$ **(2)** $$ \frac{y}{x^3} $$ **(3)** $$ x^2 $$ **(4)** $$ \frac{p^3}{27x^3} $$ **(5)** $$ -8x^2y^8 $$ **(6)** $$ -\frac{nm^3}{2} $$ **(7)** $$ 7x^2y^4 $$ **(8)** $$ 2^{-4x-2} $$ **(9)** $$ -\frac{1}{y^5} $$

Exercise 7 Set 2

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