Multiply 26244 by the smallest number so that the product is a perfect cube- What is that number- Also the cube root of the product-

Question

Toppr Admin · Accepted Answer

3-x221A-26244-3-x221A-22-xD7-38-3-x221A-22-xD7-32-xD7-36-9-x221A-22-xD7-32If 6 is a multiplied-93-x221A-23-xD7-33-54-x2234- We need 6 to be multiplied to the no- to make it a perfect cube-

Multiply $26244$ by the smallest number so that the product is a perfect cube. What is that number? Also find the cube root of the product.

$\sqrt[3]{26244} = \sqrt[3]{2^{2} \times 3^{8}} = \sqrt[3]{2^{2} \times 3^{2} \times 3^{6}} = 9 \sqrt{2^{2} \times 3^{2}}$
If $6$ is a multiplied
$= 9 \sqrt[3]{2^{3} \times 3^{3}} = 54$
$∴$ We need $6$ to be multiplied to the no. to make it a perfect cube.

Multiply 26244 by the smallest number so that the product is a perfect cube. What is that number? Also find the cube root of the product.

3√26244=3√22×38=3√22×32×36=9√22×32If 6 is a multiplied=93√23×33=54∴ We need 6 to be multiplied to the no. to make it a perfect cube.

Multiply $26244$ by the smallest number so that the product is a perfect cube. What is that number? Also find the cube root of the product.

$\sqrt[3]{26244} = \sqrt[3]{2^{2} \times 3^{8}} = \sqrt[3]{2^{2} \times 3^{2} \times 3^{6}} = 9 \sqrt{2^{2} \times 3^{2}}$
If $6$ is a multiplied
$= 9 \sqrt[3]{2^{3} \times 3^{3}} = 54$
$∴$ We need $6$ to be multiplied to the no. to make it a perfect cube.