Class-8 Cube & Cube Roots Worksheet | PDF | Mathematical Objects ... - Free Printable

Here are the step-by-step solutions for the practice questions in the image.

1. Find the cube roots of the integer:

a) $-474552$
* First, ignore the negative sign and find the cube root of $474552$.
* The number ends in $2$, so the cube root must end in $8$ (since $8^3 = 512$).
* Remove the last three digits ($552$), leaving $474$.
* Find the closest perfect cube less than $474$. We know $7^3 = 343$ and $8^3 = 512$. So, the first digit is $7$.
* The root is $78$.
* Since the original number was negative, the answer is $-78$.

b) $-2744000$
* Ignore the negative sign.
* Remove the three zeros at the end. This means the root will have one zero at the end.
* Now find the cube root of $2744$.
* It ends in $4$, so the root ends in $4$.
* Look at the remaining part: $2$. The closest cube less than or equal to $2$ is $1^3=1$. So the first digit is $1$.
* The root of $2744$ is $14$.
* Add the zero back: $140$.
* Apply the negative sign: $-140$.

2. How many digits will be there in the cube root of 46656?

* Group the digits from right to left in sets of three: $(46)(656)$.
* There are two groups.
* Therefore, the cube root will have 2 digits.
*(Check: $\sqrt[3]{46656} = 36$, which has 2 digits).*

3. Find the value of following cube roots: $\sqrt[3]{27 \times 2744}$

* Split the root: $\sqrt[3]{27} \times \sqrt[3]{2744}$.
* $\sqrt[3]{27} = 3$.
* For $\sqrt[3]{2744}$: Ends in $4$, so root ends in $4$. Remaining part is $2$, closest cube is $1^3$. So, $14$.
* Calculation: $3 \times 14 = 42$.
* Answer: $42$.

4. Find the cube root of:

a) $\frac{0.008}{0.125}$
* $\sqrt[3]{0.008} = 0.2$ (since $2^3=8$).
* $\sqrt[3]{0.125} = 0.5$ (since $5^3=125$).
* Divide: $\frac{0.2}{0.5} = \frac{2}{5} = \mathbf{0.4}$.

b) $\frac{686}{1024}$
* Simplify the fraction first by dividing numerator and denominator by $2$: $\frac{343}{512}$.
* $\sqrt[3]{343} = 7$ (since $7 \times 7 \times 7 = 343$).
* $\sqrt[3]{512} = 8$ (since $8 \times 8 \times 8 = 512$).
* Answer: $\frac{7}{8}$.

5. Prove that if a number is trebled then its cube is 27 times the cube of the given number.

* Let the number be $x$. Its cube is $x^3$.
* If the number is trebled, it becomes $3x$.
* The cube of the new number is $(3x)^3$.
* $(3x)^3 = 3^3 \times x^3 = 27x^3$.
* This shows the new cube is 27 times the original cube ($x^3$). Proved.

6. The volume of a cube is $9261000 \text{ m}^3$. Find the side of the cube.

* Side $= \sqrt[3]{\text{Volume}} = \sqrt[3]{9261000}$.
* Remove the three zeros: Root will end in $0$. Find $\sqrt[3]{9261}$.
* Ends in $1$, so root ends in $1$.
* Remaining part $9$. Closest cube is $2^3=8$. So first digit is $2$.
* Root is $21$. Add the zero: $210$.
* Answer: $210 \text{ m}$.

7. The volume of a cubical box is $474.552$ cubic meters. Find the length of each side.

* Side $= \sqrt[3]{474.552}$.
* Ignore decimal for a moment: $\sqrt[3]{474552}$.
* From Question 1a, we know $\sqrt[3]{474552} = 78$.
* The original number had 3 decimal places, so the root has 1 decimal place.
* Answer: $7.8 \text{ m}$.

8. Three numbers are to one another 2:3:4. The sum of their cubes is 0.334125. Find the numbers.

* Let numbers be $2x, 3x, 4x$.
* Sum of cubes: $(2x)^3 + (3x)^3 + (4x)^3 = 8x^3 + 27x^3 + 64x^3 = 99x^3$.
* Equation: $99x^3 = 0.334125$.
* $x^3 = \frac{0.334125}{99} = 0.003375$.
* $x = \sqrt[3]{0.003375}$.
* $\sqrt[3]{3375} = 15$ (ends in 5, $3^3=27 < 33$).
* Adjust decimal (3 places in number $\rightarrow$ 1 place in root): $x = 0.15$.
* Numbers:
* $2(0.15) = \mathbf{0.3}$
* $3(0.15) = \mathbf{0.45}$
* $4(0.15) = \mathbf{0.6}$

9. Given that $\sqrt[3]{99} = 4.626$, find the value of $\sqrt[3]{792}$.

* Notice that $792 = 99 \times 8$.
* $\sqrt[3]{792} = \sqrt[3]{99 \times 8} = \sqrt[3]{99} \times \sqrt[3]{8}$.
* Substitute values: $4.626 \times 2$.
* Answer: $9.252$.

10. Given that $\sqrt[3]{31} = 3.141$, find the value of $\sqrt[3]{\frac{248}{216}}$.

* Numerator: $248 = 31 \times 8$. So $\sqrt[3]{248} = \sqrt[3]{31} \times \sqrt[3]{8} = 3.141 \times 2 = 6.282$.
* Denominator: $\sqrt[3]{216} = 6$ (since $6^3=216$).
* Divide: $\frac{6.282}{6}$.
* Answer: $1.047$.

11. Find the volume of a cube, one face which has an area of $64 \text{ m}^2$.

* Area of one face $= \text{side}^2 = 64$.
* Side $= \sqrt{64} = 8 \text{ m}$.
* Volume $= \text{side}^3 = 8^3 = 512$.
* Answer: $512 \text{ m}^3$.

12. Find the volume of a cube whose surface area is $384 \text{ m}^2$.

* Surface Area $= 6 \times \text{side}^2$.
* $384 = 6s^2 \rightarrow s^2 = \frac{384}{6} = 64$.
* Side $s = \sqrt{64} = 8 \text{ m}$.
* Volume $= 8^3 = 512$.
* Answer: $512 \text{ m}^3$.

13. Three numbers are to one another as 2:3:4. The sum of their cubes is 33957. Find the numbers.

* Similar to Q8. Sum of cubes $= 99x^3$.
* $99x^3 = 33957$.
* $x^3 = \frac{33957}{99} = 343$.
* $x = \sqrt[3]{343} = 7$.
* Numbers:
* $2(7) = \mathbf{14}$
* $3(7) = \mathbf{21}$
* $4(7) = \mathbf{28}$

14. What is the smallest number by which 243000 must be divided so that the quotient is perfect cube?

* Prime factorization of $243000$:
* $243 = 3^5$
* $1000 = 10^3 = 2^3 \times 5^3$
* Total: $2^3 \times 3^5 \times 5^3$.
* To be a perfect cube, all powers must be multiples of 3.
* $2^3$ is good. $5^3$ is good.
* $3^5$ needs to become $3^3$. We must remove $3^2$ (which is 9).
* Answer: $9$.

15. Evaluate:

a) $\sqrt[3]{\frac{0.027}{0.008}} \div \sqrt{\frac{0.09}{0.04}} - 1$
* Cube root part: $\frac{\sqrt[3]{0.027}}{\sqrt[3]{0.008}} = \frac{0.3}{0.2} = 1.5$.
* Square root part: $\frac{\sqrt{0.09}}{\sqrt{0.04}} = \frac{0.3}{0.2} = 1.5$.
* Expression: $1.5 \div 1.5 - 1 = 1 - 1 = 0$.
* Answer: $0$.

b) $\sqrt[3]{64 \times 729}$
* $\sqrt[3]{64} = 4$.
* $\sqrt[3]{729} = 9$ (since $9^3=729$).
* $4 \times 9 = 36$.
* Answer: $36$.

c) $\sqrt[3]{1000} + \sqrt[3]{0.008} + \sqrt[3]{0.125}$
* $\sqrt[3]{1000} = 10$.
* $\sqrt[3]{0.008} = 0.2$.
* $\sqrt[3]{0.125} = 0.5$.
* Sum: $10 + 0.2 + 0.5 = 10.7$.
* Answer: $10.7$.

16. Is 53240 a perfect cube? If not, then by which smallest natural number should 53240 be divided so that the quotient is a perfect cube?

* Prime factorization of $53240$:
* $53240 = 10 \times 5324 = 2 \times 5 \times 4 \times 1331 = 2^3 \times 5 \times 11^3$.
* Powers: $2^3$ (good), $11^3$ (good), $5^1$ (bad).
* It is not a perfect cube because of the single 5.
* To make it a perfect cube, we must divide by $5$.

──────────────────────────────────────

Final Answer:
1. a) -78, b) -140
2. 2 digits
3. 42
4. a) 0.4, b) 7/8
5. Proof provided in steps.
6. 210 m
7. 7.8 m
8. 0.3, 0.45, 0.6
9. 9.252
10. 1.047
11. 512 m³
12. 512 m³
13. 14, 21, 28
14. 9
15. a) 0, b) 36, c) 10.7
16. No; divide by 5