Question

How many numbers from the following are the perfect cubes
(i) 125
(ii) 243
(iii) 343
(iv) 256
(v) 729
(vi) 1331
(vii) 8000
(viii) 9261
(ix) 5324
(x) 3375

Answer 1

concept- Perfect cube root means on $(a{)}^{\frac{1}{3}}=(a^3{)}^{\frac{1}{3}}$

And it gives natural number only not any decimal or fractional value

1. $(125{)}^{\frac{1}{3}}=(5^3{)}^{\frac{1}{3}}$

2. $(243{)}^{\frac{1}{3}}=({6.24}^3{)}^{\frac{1}{3}}$it is not perfect cube root

3.$(343{)}^{\frac{1}{3}}=(7^3{)}^{\frac{1}{3}}$

4.$(256{)}^{\frac{1}{3}}=({6.34}^3{)}^{\frac{1}{3}}$ it is not perfect cube root

5.$(729{)}^{\frac{1}{3}}=(9^3{)}^{\frac{1}{3}}$

6.$(1331{)}^{\frac{1}{3}}=({11}^3{)}^{\frac{1}{3}}$

7.$(8000{)}^{\frac{1}{3}}=({20}^3{)}^{\frac{1}{3}}$

8.$(9261{)}^{\frac{1}{3}}=({21}^3{)}^{\frac{1}{3}}$

9.$(5324{)}^{\frac{1}{3}}=({17.46}^3{)}^{\frac{1}{3}}$ it is not perfect cube root

10.$(3375{)}^{\frac{1}{3}}=({15}^3{)}^{\frac{1}{3}}$

Hence solution is $1,3,5,6,7,8,10$. Total $7$