Volume of cuboid = 3×6×9 =162 cubic units

Volume of cube = 3×3×3= 27 cubic units

Number of cubes that can be fit in the cuboid is $\frac{162}{27}$ =  6

The sum of first 5 natural numbers can be expressed as ${\left(\frac{n(n+1)}{2}\right)}^2$ where n = 5, which gives us
${\left(\frac{(5\times 6)}{2}\right)}^2$ =${\left(\frac{30}{2}\right)}^2$ = ${15}^2$
Thus the cubes of first 5 natural numbers can be expressed as the square of 15.

Cube of 1 = 1

Cube of 2 = 8

Cube of 3 = 27

Cube of 4 = 64

So total 4 numbers have their cubes that lie between 0 and 100

Cube root of 8 is 2 and cube root of -8 is -2. When they are multiplied together, the result is -4.

The cube of 3 = 27

The cube of 4 = 64

Since 30 lies between 27 & 64 so its cube root must lie between 3 & 4.
Hence, the given statement is false.

Cube of 0 is 0 (anything multiplied to 0 is 0)

8= 2 x 2 x 2

Cube root of 8 = 2

So 2+  $\frac{1}{2}$ = 2.5

By prime factorization, we get 729 = $3\times 3\times 3\times 3\times 3\times 3=3^6$
So cube root of $729=(729{)}^{\frac{1}{3}}=3^{\frac{6}{3}}=3^2=9$
Since, the cube root of 729 is a natural number, 729 is a perfect cube.

Take $a=2$ and $b=1$ and check.
Alternatively, we know the algebraic identity,

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

 Hence , we can see that $(a-b)$ is a factor of $a^3$ - $b^3$
Thus, $a^3$ - $b^3$ is divisible by $(a-b)$

For a number to be a perfect cube, its prime factors should be in the form of triplets i.e. every prime factor should be raised to the power in multiples of 3
Prime factorizing the given number, we get $360=2\times 2\times 2\times 3\times 3\times 5$
Hence, we can see that the given number must be multiplied by $3\times 5\times 5$ i.e.75