Let the side of the given cubical tank be 'a' metres .

Volume of the given cube is $a\times a\times a$

By prime factorization, we get cube root of 1331 as 11 

Surface area of each side of a cube is $a\times a$

Total surface area of the cube is $6\times a\times a$

 Cost incurred is then $10\times 6\times 11\times 11$ = ₹ 7260

Sum of cubes of first n natural numbers is $\frac{\left(n\right(n+1){)}^2}{4}$

Putting n = 7, 
$\frac{\left(7\right(7+1){)}^2}{4}$
$\frac{(7\times 8{)}^2}{4}$

= 784.

$a^3$- $b^3$ = $(a-b)$( $a^2$+$ab$+ $b^2$)

$a^3$+ $b^3$ =$(a+b)$( $a^2$-$ab$+ $b^2$)

Puting the values of a and b, we get the value of p as 351.
Alternatively, we can observe that since a and b are positive, so denominator has to be greater than numerator. Thus, $a^3$+ $b^3$ will be greater than $a^3$- $b^3$

The units digit of the cube of any number with 6 in its unit's place is always 6. Alternatively, we can see that 6 when multiplied with itself three times yields 216, whose last digit is 6. 

For every 10 mins passage of time, the number of students joining the class gets tripled i.e. multiplied by 3
The number of students joining after 40 minutes from now on is $729\times 3\times 3\times 3\times 3\times 3\times 3$ ( from 20 mins back to 40 mins forward)
The cube root of 729 is 9.
 Cube root of given number is $9\times 3\times 3=81$

Take two numbers, for example, 2 and 1 and check.

The difference of cubes of 2 and 1 is 7, which is not divisible by 3.

The units digit of the cube of 7 i.e.343 is 3.
Thus, the units digit of the cube of the number whose unit digit is 7 will be 3.

-  8  = - $2\times -2\times -2$

$\sqrt[3]{3-8}=-2$

-  216  = $-6\times -6\times -6$

$\sqrt[3]{3-216}=-6$

So the given expression $\sqrt[3]{3-8}-\sqrt[3]{3-216}$ becomes - 2 - (- 6) = - 2 + 6

= 4

For a number to be a perfect cube, its prime factors should be expressed as triplets
By prime factorization, we get 1715 = 7×7×7×5

We have three 7’s but only one 5 so we need to divide the number by 5.