Cubes and Cube Roots Test - 22

Given, to find $$\sqrt[3]{12167\times 1728}$$.

Prime factorising, we get,

$$12167= 23\times23\times23=23^3$$

and $$1728$$ $$= 2\times2\times2\times2\times2\times2\times3\times3\times3 \\ =2^3 \times 2^3 \times 3^3.$$

Then,

$$\sqrt[3]{12167\times 1728}$$

$$= \sqrt[3]{23^3}\times \sqrt[3]{2^3 \times 2^3 \times 3^3}$$.

$$= 23\times2\times2\times 3$$

$$= 276$$.

Hence, option $$A$$ is correct.

Prime factorising $$11979$$, we get,

$$ 11979 = 3 \times 3 \times 11 \times 11 \times 11$$

            $$= 3^2 \times 11 ^3$$.

We know, a perfect cube has multiples of $$3$$ as powers of prime factors.

Here, number of $$3$$'s is $$2$$ and number of $$11$$'s is $$3$$.

So we need to multiply another $$3$$ to the factorization to make $$11979$$ a perfect cube.

Hence, the smallest number by which $$11979$$ must be multiplied to obtain a perfect cube is $$3$$.

Hence, option $$D$$ is correct.

Given, the number is $$388$$.

Here, the units digit is $$8$$.

We know, the cube of $$8$$, i.e. $$8^3=512$$, whose units place is $$2$$.

Therefore, the units digit of the cube of $$388$$ is $$2$$.

Hence, option $$C$$ is correct.

Given, the number is $$109$$.

Here, the units digit is $$9$$.

We know, the cube of $$9$$, i.e. $$9^3=729$$, whose units place is $$9$$.

Therefore, the units digit of the cube of $$109$$ is $$9$$.

Hence, option $$C$$ is correct.

Given, the number is $$77774$$.

Here, the units digit is $$4$$.

We know, the cube of $$4$$, i.e. $$4^3=64$$, whose units place is $$4$$.

Therefore, the units digit of the cube of $$77774$$ is $$4$$.

Hence, option $$A$$ is correct.