2. The volume of a cubical box in 13.824 cubic metres. Find the length of
each side of the box.
3. By what number 88209 be divided so that the quotient is a perfect cube?
Find that number. Also find the cube root of the product.
4. Find the smallest number that must be subtracted from 682 to make it a
perfect cube.
5. Find the smallest numbers that must be added to 210 so that it becomes
a perfect cube.
6. Find the cube root of 864
1372
7. Find the cube root of each of the following numbers by prime
factorisation method.
a. 64
b.512 c. 10648 d. 27000
e. 15625
f. 13824 g. 110592 h.46656
Answers
Answer:
f. 13824 g. 110592 h.46656
Here is your answer mate
3:-Out of the prime factors of 88209, 11 cannot be considered in its perfect cube as it have only two factors of 11. So, 11×11 11 × 11 is the number by which 88209 must be divided to make the quotient a perfect cube. Hence, the smallest number is 121, which when divides 88209, the quotient is 729 which is a perfect cube
4:- 170
5:- 6³= 216
6:- a)864/1372. b)-343. => 6/7 Ans
7:-(i) 64
(i) 64\sqrt[3]{64}=\sqrt[3]{2\times2\times2\times2\times2\times2}
(i) 64\sqrt[3]{64}=\sqrt[3]{2\times2\times2\times2\times2\times2} 3
(i) 64\sqrt[3]{64}=\sqrt[3]{2\times2\times2\times2\times2\times2} 3
(i) 64\sqrt[3]{64}=\sqrt[3]{2\times2\times2\times2\times2\times2} 3 64
(i) 64\sqrt[3]{64}=\sqrt[3]{2\times2\times2\times2\times2\times2} 3 64
(i) 64\sqrt[3]{64}=\sqrt[3]{2\times2\times2\times2\times2\times2} 3 64 =
(i) 64\sqrt[3]{64}=\sqrt[3]{2\times2\times2\times2\times2\times2} 3 64 = 3
(i) 64\sqrt[3]{64}=\sqrt[3]{2\times2\times2\times2\times2\times2} 3 64 = 3
(i) 64\sqrt[3]{64}=\sqrt[3]{2\times2\times2\times2\times2\times2} 3 64 = 3 2×2×2×2×2×2
(i) 64\sqrt[3]{64}=\sqrt[3]{2\times2\times2\times2\times2\times2} 3 64 = 3 2×2×2×2×2×2
(i) 64\sqrt[3]{64}=\sqrt[3]{2\times2\times2\times2\times2\times2} 3 64 = 3 2×2×2×2×2×2
(i) 64\sqrt[3]{64}=\sqrt[3]{2\times2\times2\times2\times2\times2} 3 64 = 3 2×2×2×2×2×2 \sqrt[3]{64}=\ 2\times2
(i) 64\sqrt[3]{64}=\sqrt[3]{2\times2\times2\times2\times2\times2} 3 64 = 3 2×2×2×2×2×2 \sqrt[3]{64}=\ 2\times2 3
(i) 64\sqrt[3]{64}=\sqrt[3]{2\times2\times2\times2\times2\times2} 3 64 = 3 2×2×2×2×2×2 \sqrt[3]{64}=\ 2\times2 3
(i) 64\sqrt[3]{64}=\sqrt[3]{2\times2\times2\times2\times2\times2} 3 64 = 3 2×2×2×2×2×2 \sqrt[3]{64}=\ 2\times2 3 64
(i) 64\sqrt[3]{64}=\sqrt[3]{2\times2\times2\times2\times2\times2} 3 64 = 3 2×2×2×2×2×2 \sqrt[3]{64}=\ 2\times2 3 64
(i) 64\sqrt[3]{64}=\sqrt[3]{2\times2\times2\times2\times2\times2} 3 64 = 3 2×2×2×2×2×2 \sqrt[3]{64}=\ 2\times2 3 64 = 2×2
(i) 64\sqrt[3]{64}=\sqrt[3]{2\times2\times2\times2\times2\times2} 3 64 = 3 2×2×2×2×2×2 \sqrt[3]{64}=\ 2\times2 3 64 = 2×2= 4
Hope it helps you mate
