Cubes and Cube Roots Test - 26

Given, the number is $$137959^3$$.

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 $$137959$$, i.e. $$137959^3$$ is $$9$$.

Hence, option $$D$$ is correct.

On prime factorising, we get,

$$13824 = 3\times3\times3\times2\times2\times2\times2\times2\times2\times2\times2\times2$$

           $$= 2^3\times2^3\times2^3\times3^3=24^3$$.

Then, cube root of $$13824$$ is:

$$\sqrt[3]{13824}=\sqrt[3]{24^3}= 24$$.

Therefore, option $$A$$ is correct.

Prime factorising of $$4232$$ is as below,

$$\displaystyle 4232=2\times 2\times 2\times 23\times 23$$

          $$= 2 ^3 \times 23 ^2$$.

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

Here, number of $$2$$'s is $$3$$ and number of $$23$$'s is $$2$$.

So we need to multiply another $$23$$ to the factorization to make $$4232$$ a perfect cube.

Hence, the smallest number by which $$4232$$ must be multiplied to obtain a perfect cube is $$23$$.

Hence, option $$D$$ is correct.

On prime factorising, we get,

$$3375=\underline{3\times 3\times 3}\times \underline {5\times5\times5}$$ $$=3^3 \times 5^3=15^3$$.

Therefore, value of $$\sqrt[3]{3375}$$ is:

$$\sqrt[3]{3375}=\sqrt[3]{(15)^3}= 15$$.

Therefore, option $$B$$ is correct.

Step $$I$$: Form groups of $$3$$ starting from right most digit of $$343000$$, i.e. $$\displaystyle \overline { 343 }\: \overline {000} $$.

Then, the two groups are $$343$$ and $$000.$$

Here, $$343$$ has $$3$$ digit and $$000$$ has $$3$$ digit.

Step $$II$$: Take $$000$$.

Digit in unit place $$= 0$$.

Therefore, we take one's place of required cube root as $$0$$.

Step $$III$$: Now, take the other group $$343$$.

We know, $$\displaystyle {7}^{ 3 }=343$$.

Therefore, we take $$7$$ as ten's place.

$$\displaystyle \therefore \quad \sqrt[3] { 343000 } =70$$.

Hence, option $$B$$ is correct.