Exercise 6.2 (Revised) - Chapter 7 - Cube & Cube Roots - Ncert Solutions class 8 - Maths

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Chapter 6 - Cube & Cube Roots | NCERT Solutions for Class 8 Maths

Ex 6.2 Question 1.

Find the cube root of each of the following numbers by prime factorization method:
(i) 64
(ii) 512
(iii) 10648
(iv) 27000
(v) 15625
(vi) 13824
(vii) 110592
(viii) 46656
(ix) 175616
(x) 91125

Answer.

(i) 64

$
\begin{aligned}
& \sqrt[3]{64}=\sqrt[3]{2 \times 2 \times 2 \times 2 \times 2 \times 2} \\
& \sqrt[3]{64}=2 \times 2=4
\end{aligned}
$
(ii) 512

$
\begin{aligned}
& \sqrt[3]{512}=\sqrt[3]{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2} \\
& =2 \times 2 \times 2=8
\end{aligned}
$
(iii) 10648

$
\begin{aligned}
& \sqrt[3]{10648}=\sqrt[3]{2 \times 2 \times 2 \times 11 \times 11 \times 11} \\
& =2 \times 11=22
\end{aligned}
$
(iv) 27000

$
\begin{aligned}
& \sqrt[3]{27000}=\sqrt[3]{2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 5 \times 5 \times 5} \\
& =2 \times 3 \times 5=30
\end{aligned}
$
(v) 15625

$
\begin{aligned}
& \sqrt[3]{15625}=\sqrt[3]{5 \times 5 \times 5 \times 5 \times 5 \times 5} \\
& =5 \times 5=25
\end{aligned}
$
(vi) 13824

$
\begin{aligned}
& \sqrt[3]{13824}=\sqrt[3]{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3} \\
& =2 \times 2 \times 2 \times 3=24
\end{aligned}
$
(vii) 110592

$
\begin{aligned}
& \sqrt[3]{110592}=\sqrt[3]{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3} \\
& =2 \times 2 \times 2 \times 2 \times 3=48
\end{aligned}
$
(viii) 46656

$
\begin{aligned}
& \sqrt[3]{46656}=\sqrt[3]{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3} \\
& =2 \times 2 \times 3 \times 3=36
\end{aligned}
$
(ix) 175616

$
\begin{aligned}
& \sqrt[3]{175616}=\sqrt[3]{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 7 \times 7 \times 7} \\
& =2 \times 2 \times 2 \times 7=56
\end{aligned}
$
(x) 91125

Ex 6.2 Question 2.

State true or false:
(i) Cube of any odd number is even.
(ii) A perfect cube does not end with two zeroes.
(iii) If square of a number ends with 5 , then its cube ends with 25 .
(iv) There is no perfect cube which ends with 8.
(v) The cube of a two digit number may be a three digit number.

(vi) The cube of a two digit number may have seven or more digits.
(vii) The cube of a single digit number may be a single digit number.

Answer.

(i) False
Since, $1^3=1,3^3=27,5^3=125, \ldots$ are all odd.
(ii) True

Since, a perfect cube ends with three zeroes.
e.g. $10^3=1000,20^3=8000,30^3=27000, \ldots$ so on
(iii) False

Since, $5^2=25,5^3=125,15^2=225,15^3=3375$
(Did not end with 25)
(iv) False

Since $12^3=1728$
[Ends with 8]
And $22^3=10648$
[Ends with 8]
(v) False Since $10^3=1000$
[Four digit number]
And $11^3=1331$

[Four digit number]
(vi) False Since $99^3=970299$
[Six digit number]
(vii) True
$
1^3=1
$
[Single digit number]
$
2^3=8
$
[Single digit number]