Question

please answer this question​

Answer 1

Find the prime factors of 2197:

$2197 = 13 \times 13 \times 13$

$2197 = 13^3$

Find the prime factors of  1728:

$1728 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3$

$1728 = 2^6 \times 3^3$

Write the numbers as a product of its prime factors:

$\sqrt[3]{2197} - \sqrt[3]{1728} = \sqrt[3]{13^3} - \sqrt[3]{2^6 \times 3^3}$

Solve:

$\sqrt[3]{2197} - \sqrt[3]{1728} = \sqrt[3]{13^3} - \sqrt[3]{(2^6 \times 3^3)}$

$\sqrt[3]{2197} - \sqrt[3]{1728} = \sqrt[3]{13^3} - \sqrt[3]{(2^2 \times 3)^3}$

$\sqrt[3]{2197} - \sqrt[3]{1728} =13- (2^2 \times 3)$

$\sqrt[3]{2197} - \sqrt[3]{1728} =13- 12$

$\sqrt[3]{2197} - \sqrt[3]{1728} = 1$

Answer: (D) 1

Answer 2

Answer:

1

Step-by-step explanation:

Given : ${\sf{ {\sqrt[3]{2197}} - {\sqrt[3]{1728}}}}$

• When we remove cube root from a number, then the number is raised to the power of 1/3.

$\implies{\sf{ (2197)^{ {\frac{1}{3}} } - (1728)^{ {\frac{1}{3}} } }}$

• Cube of 13 = (13)³ = 2197
• Cube of 12 = (12)³ = 1728

$\implies{\sf{ (13^3)^{ {\frac{1}{3}} } - (12^3)^{ {\frac{1}{3}} } }}$

${\boxed{\tt{\bigstar \ \ Identity \ : \ (a^m)^n = a^{m \times n}}}}$

$\implies{\sf{ (13)^{3 \times {\frac{1}{3}} } - (12)^{3 \times {\frac{1}{3}} }}}$

$\implies{\sf{ (13)^1 - (12)^1}}$

$\implies{\sf{ 13 - 12}}$

$\implies{\boxed{\boxed{\sf{1}}}}$

Hence, the answer is (D) 1.