Question

find the smallest number by which 59400 must be multiplied to make the product a perfect cube. also find the cube root of the product.​ hoe to solve this step by step plzz dont spam plzz​

Answer 1

Prime factorising 68600, we get,

68600=2×2×2×5×5×7×7×7

=2

3

×5

2

×7

3

.

We know, a perfect cube has multiples of 3 as powers of prime factors.

Here, number of 2's is 3, number of 5's is 2 and number of 7's is 3.

So we need to multiply another 5 in the factorization to make 68600 a perfect cube.

Hence, the smallest number by which 68600 must be multiplied to obtain a perfect cube is 5.

hope it helps u big bro

Answer 2

Answer:

Given :

59400

To find :

Smallest Number by which given number must be multiplied to make the product a perfect cube.

Cube root of product (new Number)

Method used :

First of all find prime factor of given number

Then make group of prime factor such that each group contain Same number 3 time { example (2×2×2) }

Then multiply the given number by Number whose pairs can't be formed { like 5 } , so to make a group of 3

Solution :

Part 1 :

Prime factor -

59400 = 2 × 2 × 2 × 3 × 3 × 3 × 5 × 5 × 11

59400 = (2×2×2) × (3×3×3) × (5×5) × (11)

So , we should multiply by , (5 × 11 × 11 )

New Number = 59400 × 605

New number = 35937000

Part 2 :

Cube root of new Number

35937000 = (2×2×2) × (3×3×3) × (5×5×5) × (11×11×11)

$\sqrt[3]{35937000} = \sqrt[3]{(2*2*2) (3*3*3) (5*5*5) (11*11*11)}$

$\sqrt[3]{35937000} = \sqrt[2]{2^{3} * 3^{3} * 5^{3} * 11^{3} }$

$\sqrt[3]{35937000} = \sqrt[3]{(2*3*5*11)^{3} }$

$\sqrt[3]{35937000} = 2*3*5*11$

$\sqrt[3]{35937000} = 330$

ANSWER :

605

330

Explanation: