Let's go through both questions step by step.
Question 52:
We are given the expression [tex]$\sqrt[3]{a^3} \times \sqrt[3]{-b^6}$[/tex] and need to simplify it.
1. First, consider [tex]$\sqrt[3]{a^3}$[/tex]. The cube root of [tex]$a^3$[/tex] is [tex]$a$[/tex], because:
[tex]\[
\sqrt[3]{a^3} = a
\][/tex]
2. Now, consider [tex]$\sqrt[3]{-b^6}$[/tex]. The term [tex]$-b^6$[/tex] can be broken down as [tex]$(-1) \times (b^6)$[/tex]. The cube root of [tex]$b^6$[/tex] is [tex]$b^2$[/tex] (since [tex]$(b^2)^3 = b^6$[/tex]) and the cube root of [tex]$-1$[/tex] is [tex]$-1$[/tex]. Therefore:
[tex]\[
\sqrt[3]{-b^6} = \sqrt[3]{-1 \times b^6} = \sqrt[3]{-1} \times \sqrt[3]{b^6} = -1 \times b^2 = -b^2
\][/tex]
3. Combining these results, we have:
[tex]\[
\sqrt[3]{a^3} \times \sqrt[3]{-b^6} = a \times (-b^2) = -a b^2
\][/tex]
So, the value of [tex]$\sqrt[3]{a^3} \times \sqrt[3]{-b^6}$[/tex] is [tex]$-a b^2$[/tex]. Therefore, the correct option is:
[tex]\[
\text{(3) } -a b^2
\][/tex]
Question 53:
We need to find the numbers whose cube and cube root both are equal.
1. The cube and cube root of a number [tex]\( n \)[/tex] being the same means:
[tex]\[
n^3 = n \quad \text{and} \quad \sqrt[3]{n} = n
\][/tex]
2. Let's test some simple numbers:
- For [tex]\( n = 1 \)[/tex]:
[tex]\[
1^3 = 1 \quad \text{and} \quad \sqrt[3]{1} = 1
\][/tex]
Both conditions are satisfied for [tex]\( n = 1 \)[/tex].
- For [tex]\( n = -1 \)[/tex]:
[tex]\[
(-1)^3 = -1 \quad \text{and} \quad \sqrt[3]{-1} = -1
\][/tex]
Both conditions are satisfied for [tex]\( n = -1 \)[/tex].
3. Therefore, the numbers that satisfy these conditions are 1 and -1.
So, the numbers whose cube and cube root both are equal are 1 and -1. Therefore, the correct option is:
[tex]\[
\text{(3) Both (1) \& (2)}
\][/tex]
In conclusion:
- For Question 52: The answer is [tex]\((3) -a b^2\)[/tex].
- For Question 53: The answer is [tex]\((3) Both (1) \& (2)\)[/tex].
