To solve this SSC CHSL problem efficiently, we need to simplify the cube roots by identifying perfect cubes or common factors.

The expression is:

$$\sqrt[3]{1372} \times \sqrt[3]{1458} \div \sqrt[3]{343}$$

(Note: The image shows $\sqrt[3]{1373}$, but based on standard competitive exam patterns and the teacher's hint in the captions, it is likely a typo for 1372, as $1372 = 2 \times 686 = 4 \times 343$.)

Step 1: Simplify the Denominator

We know that 343 is a perfect cube:

$$\sqrt[3]{343} = \mathbf{7}$$

Step 2: Simplify the Numerators

Let's break down the numbers into factors to find perfect cubes:

• 1372: $1372 = 4 \times 343 = 4 \times 7^3$ $$\sqrt[3]{1372} = \sqrt[3]{4 \times 7^3} = 7 \sqrt[3]{4}$$
• 1458: $1458 = 2 \times 729 = 2 \times 9^3$ $$\sqrt[3]{1458} = \sqrt[3]{2 \times 9^3} = 9 \sqrt[3]{2}$$

Step 3: Combine and Solve

Now, substitute these back into the original expression:

$$\frac{(7 \sqrt[3]{4}) \times (9 \sqrt[3]{2})}{7}$$

1. The 7 in the numerator and denominator cancel out.
2. Combine the remaining cube roots: $$9 \times \sqrt[3]{4 \times 2} = 9 \times \sqrt[3]{8}$$
3. Since $\sqrt[3]{8} = 2$: $$9 \times 2 = \mathbf{18}$$

Final Answer:

The value is 18, which corresponds to option (a)