Cubes and Cube Roots Test - 27

On prime factorising, we get,

$$27=\underline{3\times 3\times3}$$ $$=3^3$$.

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.

$$\displaystyle \therefore \quad \sqrt[3] { 4913 } =17$$

Hence, option $$D$$ is correct.

By prime factorising $$1352$$, 

we get, $$ 1352 = 2 \times 2 \times 2 \times 13 \times 13$$

                        $$= 2 ^3 \times 13 ^2$$

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

Here, number of $$2$$'s is $$3$$ and number of $$13$$'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,

$$ 216=\underline { 2 \times 2 \times 2 } \times \underline { 3 \times 3 \times 3 }=2^3\times3^3=6^3$$.

$$ \Rightarrow { -216 } =(-6)^3 $$.

On prime factorising, we get,

$$64=\underline{4\times 4\times4}$$ $$=4^3$$.

Then, $$-64$$ $$=(-4)^3$$.

Therefore, value of $$\sqrt[3]{64}\div\sqrt[3]{-64}$$ is:

$$\sqrt[3]{64}\div\sqrt[3]{-64}$$

$$=\sqrt[3]{4^3}\div\sqrt[3]{(-4)^3}$$

$$=4\div(-4)=-1$$.

Therefore, option $$A$$ is correct.