Question

0. What is the smallest number by which 4116 must be multiplied so that the product is a
perfect cube? Find the cube root of the perfect cube so obtained.
htained​

Answer 1

Step-by-step explanation:

First you have to do prime factorization

so I got answer

=2×2×3×(7×7×7 )

so here we got a triplet

to make 4116 as a perfect cube you have to multipy with 18 because

18 = 2×3×3 by multiplying 18 you will get perfect cube

4116×18 = 74088

74088= (2×2×2)(3×3×3)(7×7×7)

= consider only one of them

=2×3×7

= 42

so 42³=74088

Answer 2

Step 1: Do prime factorization to find the product of primes.

$\begin{array}{r | l} 2 & 4116 \\ \cline{2-2} 2 & 2058 \\ \cline{2-2} 3 & 1029 \\ \cline{2-2} 7 & 343 \\ \cline{2-2} 7 & 49 \\ \cline{2-2} & 7 \\ \end{array}$

4116 ⇒ 2×2×3×7×7×7

Step 2: Group the product of primes with 3 numbers in each.

$\sf 2\times 2\times (7\times 7 \times 7) \times 3$

∴4116 is not a perfect square.

∴ We must multiply 18 (2×3×3) to make 4116 a perfect cube.

Step 3: Multiply 9 to 4116

4116×18 = 37044

Step 4: Do prime factorization for 37044 to obtain the product of primes.

$\begin{array}{r | l} 2 & 74088\\ \cline{2-2} 2 & 37044 \\ \cline{2-2} 2 & 18522 \\ \cline{2-2} 3 & 9261 \\ \cline{2-2} 3 & 3087 \\ \cline{2-2} 3 & 1029 \\ \cline{2-2} 7 &343 \\ \cline{2-2} 7 & 49 \\\cline{2-2} & 7 \\ \end{array}$

Step 5: Group the product of primes with 3 numbers in each.

$\sf (2\times2\times 2)\times (3\times 3\times 3)\times (7\times 7\times 7)$

Step 6: To obtain the perfect square we need to take one number from each pair and multiply.

$\sf \sqrt{74088} = 2\times 3\times 7$

$\sf \sqrt{74088} =42$

∴The smallest number which 4116 must be multiplied to make it a perfect square is 18. The cube root found after multiplying 18 to 4116 would be 42.