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Find the cube roots by prime factorization method

(i) 15625

(ii) 2744

(iii) 125/2197

(iv) 5832

(v) 64000

(vi) 10648

(vii) 1331

(viii) 1728

(i) $15625 = 5 \times 5 \times 5 \times 5 \times 5 \times 5 $

$\sqrt[3]{15625} =5$

(ii) $2744= 2^3 \times 7^3$

$\sqrt[3]{2744} =2 \times 7=14$

(iii) $125/2197= (5 x 5 x 5) / (13 \times 13 \times 13)$

$\sqrt[3]{125/2197} =5 /13$

(iv) 5832= 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 3

$\sqrt[3]{5832} =2 \times 3 \times 3=18$

(v)64000 = 4 x 4 x 4 x 10 x 10 x 10

$\sqrt[3]{64000} =4 \times 10 =40$

(vi) 10648= 2 x 2 x 2 x 11 x 11 x 11

$\sqrt[3]{10648} =2 \times 11 =22$

(vii) 1331=11 x 11 x 11

$\sqrt[3]{1331} =11$

(viii) 1728= 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3

$\sqrt[3]{1728} =12$

Find the cubes of these numbers.

(i) .4

(ii) 8

(iii)15

(iv).02

(v) 1.1

(vi).011

(i) $.4^3 = .046 $

(ii) $8^3= 512$

(iii) $15^3=3375$

(iv) $.02^3= .000008$

(v)$1.1^3= 1.331$

(vi)$.011^3= .000001331$

Is 256 a perfect cube? If not, find the smallest number by which it should be divided to get a perfect cube.

$256= 2^8= 2^3 \times 2^3 \times 2^2$

Therefore it is not a perfect cube

So, it must be divided by 4 to become perfect cube

if $6^{n+2}=1296$, then $\sqrt[3]{n + 727}$ is

$6^{n+2}=1296$

$6^{n+2}=6^4$

$n+2=4$ or $n=2$

Now $\sqrt[3]{n + 727}= \sqrt[3]{2 + 727}=\sqrt[3]{729}=9$

Three numbers are in the ratio 1:2:3 and the sum of their cubes is 4500. Find the numbers.

Let the number be x,2x,3x

Now

$x^3 + 8x^3 + 27x^3=4500$

$36x^3=4500$ or $x^3=125$

$x=5$

So numbers are 5,10,15

If $\sqrt[3]{157464}=54$, then find the value

$\sqrt[3]{157.464}+ \sqrt[3]{.157464}+ \sqrt[3]{.000157464}$

$\sqrt[3]{157.464}+ \sqrt[3]{.157464}+ \sqrt[3]{.000157464}=5.4 + .54 + .054=5.994$

(i)The least number by which 72 be multiplied to make it a perfect cube is _____________

(ii)If $8x^3=216$, then x ix ____

(iii) The cube of .5 is _____

(iv) There are _________ perfect cubes between 1 and 1000.

(v) The digit at the ones place of $23^3$ is ____

(vi) The cube of the even natural number is ____

(i)3

(ii)3

(iii).125

(iv) 8

(v)7

(vi) even

(i)648 is not perfect cube?

(ii) 999 is a perfect cube

(iii)Square of a number is positive, so the cube of that number will also be positive

(iv) $125 \times 8 \times 27$ is a perfect cube

(v) For an integer p, $p^3$ is always greater than $p^2$

(vi)$.8^3=5.12$

(i)True

(ii) false

(iii) false

(iv) True

(v)False

(vi) False

Cube root of the expression $125^2$

(a)5

(b)25

(c)10

(d)125

which of these is a perfect cube

(a) 216

(b) 392

(b) 8640

(d) 243

which of these is not a perfect cube

(a)729

(b)2197

(c)-1331

(d)169

$\sqrt [3]{.027} $ is

(a).3

(b) .03

(c).94

(d).33

$\sqrt [3]{216 \times 27} $ is

(a)12

(b)18

(c)21

(d)36

If m is the cube root of n, then n is

(a)$\sqrt [3]{m} $

(b) 3m

(c) m/3

(d) $m^3$

$(1 \frac {3}{10})^3$ is

(a) $1 \frac {3}{1000}$

(b) $2 \frac {27}{1000}$

(c) $3 \frac {9}{1000}$

(d) $2 \frac {197}{1000}$

The volume of the cube is $5832m^3$ , the side is

(a) 18m

(b) 16 m

(c) 28 m

(d) 19m

$\sqrt [3]{\frac {8}{27}} + \sqrt [3]{\frac {1728}{1331}} $

(a) 22/33

(b) 16/11

(c) 11/16

(d) 11/3

which of the following expression is not a perfect cube

(a) 27 x 125 x 64

(b) 1331 x 125 x 8

(c) 15 x 8 x 25 x 9

(d) None of these

The surface area of a cube whose volume is $8m^3$

(a) $32m^2$

(b) $24m^2$

(c) $12m^2$

(d) None of these

A cuboid of dimensions 16m, 4m, 125m has been melted to form a cube. Find the side of the cube

(a) $20$m

(b) $16$m

(c) $5$m

(d) None of these

9. (b)

$125^2= 125 \times 125$

So $\sqrt[3]{125^2}=\sqrt[3]{15 \times 125}=5 \times =25$

10. (a)

11. (d)

12. (a)

13. (b)

14. (d)

15 .(d)

16. (a)

17. (b)

$\sqrt [3]{\frac {8}{27}} + \sqrt [3]{\frac {1728}{1331}} = \frac {2}{3} + \frac {12}{11} = \frac {48}{33} = \frac {16}{11}$

18. (d)

19. (b)

$a^3=8$ or $a=2$

Surface area=$6a^2=24$

20. (a)

$a^3=16 \times 4 \times 125$

$a=20$

(p) -> (iii)

(q) -> (iv)

(r) -> (v)

(s) -> (ii)

(t) -> (i)

This Cube and cube roots class 8 worksheet is prepared keeping in mind the latest syllabus of CBSE . This has been designed in a way to improve the academic performance of the students. If you find mistakes , please do provide the feebback on the mail.

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