Free Cubes and Cube Roots 01 Practice Test - 8th Grade 

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.

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

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

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 $9\times 3\times 3=81$

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.

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. 

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

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

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 $a^3$- $b^3$

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.