Question

$1.\sqrt[3]{3375} \\ 2. \sqrt[3]{1728}$
find both by using prime factorization

2nd question

2 . given that the prime factorization of 9801 is
${3 }^{4} \times {11}^{2}$
,find
$\sqrt{9801}$
with working

now 3rd

3.given that the prime factorization of 21 952 is
${2}^{6} \times {7}^{3}$
find
$\sqrt[3]{21952}$
with working

​

Answer 1

=> ∛3375

Prime factorisation of 3375 = 5 × 5 × 5 × 3 × 3 × 3

=> (5)³ × (3)³  

=> 5 × 3

=> 15

Hence,

Cube root of 3375 = 15.

 ³√3375 = 15

=> ∛1728

Step 1: Find the prime factors of 1728

1728 = 2x2x2x2x2x2x3x3x3

Step 2: Group the factors in a pair of three and write in the form of cubes.

1728 = (2x2x2)x(2x2x2)x(3x3x3)

1728 = 2^3x2^3x3^3

Step 3: Apply cube root both the sides and take out the terms in cubesout of the cube root.

3√1728 = 3√(2^3x2^3x3^3) = 2 x 2 x 3 = 12

Hence, 3√1728 =12

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=>  given that the prime factorization of 9801 is 3^4 × 11^2,find√9801 with working.

We are supposed to find the square root of 9801 by prime factorization

So,

3 | 9801

3 | 3267

3 | 1089

3 | 363

11 | 121

11 |  11

 |  1

$9801=3*3*3*3*11*11\\\\\sqrt{9801} = \sqrt{3*3*3*3*11*11} \\\\\sqrt{9801}=\sqrt{3*3*11} \\\\\sqrt{9801}=99$

Hence, the square root is 99.

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=> given that the prime factorization of 21 952 is 2^6×7^3 find, ∛21952 with working.

2*2*2*2*2*2*7*7*7 is prime factorization of 21952 . cube root of the required is 2*2*7= 28