CBSE Questions for Class 8 Maths Cubes And Cube Roots Quiz 4 - MCQExams.com

Prime factorising $$25$$, we get,

$$25=5\times 5=5^2$$.

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

Here, number of $$5$$'s is $$2$$.

So we need to multiply another $$5$$ in the factorization to make $$25$$ a perfect cube.

Hence, the smallest number by which $$25$$ must be multiplied to obtain a perfect cube is $$5$$.

Prime factorising $$1296$$, we get,

$$1296= 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3$$

          $$= 2^4 \times 3 ^4$$.

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

Here, number of $$2$$'s is $$4$$ and number of $$3$$'s is $$4$$.

So we need to divide $$2$$ and $$3$$ from the factorization to make $$1296$$ a perfect cube.

Hence, the smallest number by which $$1296$$ must be divided to obtain a perfect cube is $$2 \times 3=6$$.

Prime factorising $$4096$$, we get,

$$4096 = 2\times2\times2\times2\times2\times2\times$$ $$2\times2\times2\times2\times2\times2$$ $$ = 2^{12}$$.

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

Here, number of $$2$$'s is $$12$$, which is a multiple of $$3$$.

Therefore, $$4096$$ is a perfect cube.

On prime factorisation of $$175616$$, we get,

$$ 175616= 2\times2\times2\times2\times2\times2\times2\times2\times2\times7\times7\times7$$

              $$  = 8\times8\times8\times7\times7\times7$$

              $$= 8^3 \times 7^3$$.

Then, cube root of $$175616$$ is:

$$\sqrt[3]{175616}=\sqrt[3]{8^3 \times 7^3}$$

                 $$= 8\times 7\times$$

                 $$=56$$

Therefore, option $$A$$ is correct.

Prime factorising $$1188$$, we get,

$$1188 = 2 \times 2 \times 3 \times 3 \times 3 \times 11$$

          $$= 2^2 \times 3^3 \times 11 $$.

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

Here, number of $$2$$'s is $$2$$, number of $$3$$'s is $$3$$ and number of $$11$$'s is $$1$$.

So we need to divide $$2^2$$ and $$11$$ from the factorization to make $$1188$$ a perfect cube.

Hence, the smallest number by which $$1188$$ must be divided to obtain a perfect cube is $$2^2 \times 11 = 44$$.

Prime factorising $$4320$$, we get,

$$ 4320 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 5$$

          $$= 2 ^5 \times 3 ^3 \times 5^1$$.

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

Here, number of $$2$$'s is $$5$$, number of $$3$$'s is $$3$$ and number of $$5$$'s is $$1$$.

So we need to multiply another $$2$$, and $$5^2$$ in the factorization to make $$4320$$ a perfect cube.

Hence, the smallest number by which $$4320$$ must be multiplied to obtain a perfect cube is $$2\times 5^2=50$$.

Hence, option $$C$$ is correct.

On prime factorising, we get,

$$27000 = 2\times2\times2\times5\times5\times5\times3\times3\times3$$

           $$= 2^3\times5^3\times3^3$$.

Then, cube root of $$27000$$ is:

$$\sqrt[3]{27000}=\sqrt[3]{3^3 \times 5^3 \times 2^3}= 3\times5\times2=30$$.

Therefore, $$30$$ is the required solution.

Hence, option $$A$$ is correct.

On prime factorising, we get,

$$91125=(5 \times 5\times 5)\times (3\times3\times3)\times(3\times3\times3)$$

            $$= 5^3 \times 3^3 \times 3^3$$.

Then, cube root of $$91125$$ is:

$$\sqrt[3]{91125}=\sqrt[3]{5^3 \times 3^3 \times 3^3 }=5\times 3\times 3=45$$.

Therefore, option $$A$$ is correct.

On prime factorisation of the numbers individually, we get,

$$125=\underline { 5 \times 5 \times 5 }=5^3$$.

$$1331=\underline { 11 \times 11 \times 11 }=11^3$$.

Then, cube root of $$-\dfrac{125}{1331}$$ is:

$$ \displaystyle \sqrt [ 3 ]{-\dfrac{125}{1331}}$$ $$ =\displaystyle \sqrt [ 3 ]{-\dfrac{5^3}{11^3}}$$ $$ =\displaystyle \sqrt [ 3 ]{(-\dfrac{5}{11})^3}$$ $$= \displaystyle {-\dfrac{5}{11}}$$.

Thus, option $$B$$ is correct.

On prime factorisation of the numbers individually, we get,

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

$$125=\underline { 5 \times 5 \times 5 }=5^3$$.

Therefore, cube root of $$\dfrac{27}{125}$$ is:

$$ \displaystyle {\sqrt [ 3 ]{ \dfrac{27}{125}}=\sqrt [ 3 ]{ \dfrac{3^3}{5^3}}=\dfrac{3}{5}} $$.

Therefore, option $$A$$ is correct.

On prime factorising, we get,

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

$$343=\underline{7\times 7\times 7}$$ $$=7^3$$.

Then, $$-343$$ $$=(-7)^3$$.

Therefore, value of $$\sqrt[3]{216\times(-343)}$$ is:

$$\sqrt[3]{6^3\times(-7)^3}=6\times(-7)=-42$$.

Therefore, option $$B$$ 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.

On prime factorising, we get,

$$35937=(3\times3\times3)\times(11\times11\times11)$$

                  $$= 3^3 \times 11^3$$.

Then, cube root of $$35937$$ is:

$$\sqrt[3]{35937}=\sqrt[3]{3^3 \times 11^3}$$

$$= 3\times 11$$

$$=33$$.

Therefore, option $$A$$ is correct.

On prime factorising, we get,

$$1331=\underline{11\times 11\times 11}$$ $$=11^3$$.

Therefore, value of $$\sqrt[3]{1331}$$ is:

$$\sqrt[3]{1331}=\sqrt[3]{(11)^3}= 11$$.

Therefore, option $$A$$ is correct.

On prime factorising, we get,

$$1331=\underline{11\times 11\times 11}$$ $$=11^3$$.

Then, $$-1331$$ $$=(-11)^3$$.

Therefore, value of $$\sqrt[3]{-1331}$$ is:

$$\sqrt[3]{-1331}=\sqrt[3]{(-11)^3}= -11$$.

Therefore, option $$B$$ is correct.

On prime factorising, we get,

$$8000=\underline{2\times 2\times 2}\times \underline{2\times2\times2} \times \underline{5\times5\times5}$$ $$=2^3\times 2^3 \times 5^3=20^3$$.

Then, $$-8000$$ $$=(-20)^3$$.

Therefore, value of $$\sqrt[3]{-8000}$$ is:

$$\sqrt[3]{-8000}=\sqrt[3]{(-20)^3}= -20$$.

Therefore, option $$B$$ is correct.

On prime factorising, we get,

$$27000=\underline{3\times 3\times 3}\times \underline{2\times2\times2}\times \underline{5\times5\times5}$$ $$=2^3\times3^3\times5^3=30^3$$.

Then, $$-27000$$ $$=(-30)^3$$.

Therefore, value of $$\sqrt[3]{-27000}$$ is:

$$\sqrt[3]{-27000}=\sqrt[3]{(-30)^3}= -30$$.

Therefore, option $$B$$ is correct.

Prime factorising $$6912$$, we get,

$$6912 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3$$

          $$= 2 ^8 \times 3 ^3$$.

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

Here, number of $$2$$'s is $$8$$ and number of $$3$$'s is $$3$$.

So we need to multiply another $$2$$ to the factorization to make $$6912$$ a perfect cube.

Hence, the smallest number by which $$6912$$ must be multiplied to obtain a perfect cube is $$2$$.

Hence, option $$A$$ is correct.