Explanation
$$ 5400 = 3 \times 3 \times 3 \times 5 \times 5 \times 2 \times 2 \times 2$$
$$= 2 ^3 \times 3 ^3 \times 5^2$$.
We know, a perfect cube has multiples of $$3$$ as powers of prime factors.
Here, number of $$2$$'s is $$3$$, number of $$3$$'s is $$3$$ and number of $$5$$'s is $$2$$.
So we need to multiply another $$5$$ to the factorization to make $$5400$$ a perfect cube.
Hence, the smallest number by which $$5400$$ must be multiplied to obtain a perfect cube is $$5$$.
Therefore, option $$C$$ is correct.
Prime factorising $$500$$, we get,
$$ 500 = 5 \times 5 \times 5 \times 2 \times 2$$
$$= 5 ^3 \times 2^2 $$.
Here, number of $$2$$'s is $$2$$ and number of $$5$$'s is $$3$$.
So we need to multiply another $$2$$ to the factorization to make $$500$$ a perfect cube.
Hence, the smallest number by which $$500$$ must be multiplied to obtain a perfect cube is $$ 2$$.
On prime factorising, we get,
$$166375= 5\times5\times5\times11\times11\times11$$
$$= 5^3\times11^3$$.
Then, cube root of $$166375$$ is:
$$\sqrt[3]{166375}=\sqrt[3]{5^3 \times 11^3}=5\times11=55$$.
Therefore, option $$B$$ is correct.
Given, the units digit of the number is $$7$$.
We know, the cube of $$7$$, i.e. $$7^3=343$$, whose units place is $$3$$.
Therefore, the units digit of the cube is $$3$$.
Hence, option $$B$$ is correct.
Step $$I$$: Form groups of $$3$$ starting from right most digit of $$59319$$, i.e. $$\displaystyle \overline { 59 }\: \overline { 319 } $$.
Then, the two groups are $$59$$ and $$319.$$
Here, $$59$$ has $$2$$ digit and $$319$$ has $$3$$ digit.
Step $$II$$: Take $$319$$.
Digit in unit place $$= 9$$.
Therefore, we take one's place of required cube root as $$9$$ ....[Since, $$9^3=729$$].
Step $$III$$: Now, take the other group $$59$$.
We know, $$\displaystyle {3}^{ 3 }=27 $$ and $${4 }^{ 3 }=64$$.
Here, the smallest number among $$3$$ and $$4$$ is $$3$$.
Therefore, we take $$3$$ as ten's place.
$$\displaystyle \therefore \quad \sqrt[3] { 59319 } =39$$.
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.
Prime factorising of $$4232$$ is as below,
$$\displaystyle 4232=2\times 2\times 2\times 23\times 23$$
$$= 2 ^3 \times 23 ^2$$.
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 $$23$$.
Here, number of $$7$$'s is $$2$$, number of $$13$$'s is $$1$$ and number of $$19$$'s is $$3$$.
So we need to multiply another $$7$$ and $$13^2$$ to the factorization to make $$x$$ a perfect cube.
Hence, the smallest number by which $$x$$ must be multiplied to obtain a perfect cube is $$7\times 13^2=1183$$.
Therefore, option $$D$$ is correct.
$$3375=\underline{3\times 3\times 3}\times \underline {5\times5\times5}$$ $$=3^3 \times 5^3=15^3$$.
Therefore, value of $$\sqrt[3]{3375}$$ is:
$$\sqrt[3]{3375}=\sqrt[3]{(15)^3}= 15$$.
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.
$$\displaystyle \therefore \quad \sqrt[3] { 343000 } =70$$.
$$27=\underline{3\times 3\times3}$$ $$=3^3$$.
Then, $$-27$$ $$=(-3)^3$$.
Therefore, value of $$\sqrt[3]{27}\times\sqrt[3]{-27}$$ is:
$$\sqrt[3]{27}\times\sqrt[3]{-27}$$
$$=\sqrt[3]{3^3}\times\sqrt[3]{(-3)^3}$$
$$=3\times(-3)=-9$$.
Therefore, option $$A$$ is correct.
Prime factorising $$231525$$, we get,
$$231525= 3 \times 3 \times 3 \times 5 \times 5 \times 7 \times 7 \times 7$$
$$= 3^3 \times 5^2 \times 7^3$$.
We know that a perfect cube number has its factor with an exponent as multiples of $$3.$$
Here, exponent of $$3$$'s is $$3$$, exponent of $$5$$'s is $$2$$ and exponent of $$7$$'s is $$3$$.
So we need to multiply another $$5$$ to the factorization to make $$2,31,525$$ a perfect cube.
Hence, the smallest number by which $$2,31,525$$ must be multiplied to obtain a perfect cube is $$5$$.
Hence, option $$A$$ is correct.
Step $$I$$ : Form groups of $$3$$ starting from right most digit of $$4913$$, i.e. $$\displaystyle \overline { 4 }\: \overline { 913 } $$
Then, the two groups are $$4$$ and $$913.$$
Here, $$4$$ has only $$1$$ digit and $$913$$ has $$3$$ digits.
Step $$II$$ : Take $$913$$
Digit in unit place $$= 3$$
Therefore, we take one's place of required cube root as $$7$$ .... [Since, $$7^3=343$$]
Step $$III$$ : Now, take the other group $$4$$
We search for the largest cube number which is less than the number in the second group.
Number in the second group is $$4$$
And also, we know, $$1 \leq 4 < 8$$
$$\Rightarrow 1^3 \leq 4 < 2^3$$
Here, the smallest number among $$1$$ and $$2$$ is $$1$$
Therefore, we take $$1$$ as ten's place.
$$\displaystyle \therefore \quad \sqrt[3] { 4913 } =17$$
By prime factorising $$1352$$,
we get, $$ 1352 = 2 \times 2 \times 2 \times 13 \times 13$$
$$= 2 ^3 \times 13 ^2$$
Here, number of $$2$$'s is $$3$$ and number of $$13$$'s is $$2$$.
So we need to multiply another $$13$$ to the factorization to make $$1352$$ a perfect cube.
Hence, the smallest number by which $$1352$$ must be multiplied to obtain a perfect cube is $$13$$.
On prime factorisation of the numbers individually, we get,
$$ 216=\underline { 2 \times 2 \times 2 } \times \underline { 3 \times 3 \times 3 }=2^3\times3^3=6^3$$.
$$ \Rightarrow { -216 } =(-6)^3 $$.
$$ \Rightarrow { -8 } =(-2)^3$$.
Therefore,
$$\sqrt[3]{-8}-\sqrt[3]{-216}$$ $$=\sqrt[3]{(-2)^3}-\sqrt[3]{(-6)^3}=(-2)-(-6)=-2+6=4$$.
Thus, option $$C$$ is correct.
$$64=\underline{4\times 4\times4}$$ $$=4^3$$.
Then, $$-64$$ $$=(-4)^3$$.
Therefore, value of $$\sqrt[3]{64}\div\sqrt[3]{-64}$$ is:
$$\sqrt[3]{64}\div\sqrt[3]{-64}$$
$$=\sqrt[3]{4^3}\div\sqrt[3]{(-4)^3}$$
$$=4\div(-4)=-1$$.
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