CBSE Questions for Class 8 Maths Cubes And Cube Roots Quiz 5 - MCQExams.com

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.

On prime factorisation of the numbers individually, we get,

$$64=\underline { 2 \times 2 \times 2 } \times \underline { 2 \times 2 \times 2 }$$ $$ =2^3 \times 2^3=4^3$$.

$$729=\underline { 3 \times 3 \times 3 } \times \underline { 3 \times 3\times 3 } $$ $$=3^3 \times 3^3=9^3$$.

Therefore,

$$ \sqrt [ 3 ]{ 64 \times 729 }$$ $$= \sqrt [ 3 ]{4^3 \times 9^3 }$$ $$=4\times9=36$$.

Hence, option $$C$$ is correct.

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.

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.

On prime factorising, we get,

$$274625000=(5 \times 5\times 5)\times (5\times5\times5)\times(2\times2\times2)\times(13\times13\times13)$$

                  $$= 5^3 \times 5^3 \times 2^3 \times 13^3$$.

Then, cube root of $$274625000$$ is:

$$\sqrt[3]{274625000}=\sqrt[3]{5^3 \times 5^3 \times 2^3 \times 13^3}= 5\times 5\times 2\times 13=650$$.

Therefore, option $$A$$ is correct.

On prime factorising, we get,

$$614125 = 5\times5\times5\times17\times17\times17$$

              $$= 5^3\times17^3$$.

Then, cube root of $$614125$$ is:

$$\sqrt[3]{614125}=\sqrt[3]{5^3 \times 17^3}= 5\times17=85$$.

Therefore, option $$C$$ is correct.

On prime factorising, we get,

$$571787=\underline{83\times 83\times 83}$$ $$=83^3$$.

Then, $$-571787$$ $$=(-83)^3$$.

Therefore, cube root of $${-571787}$$ is:

$$\sqrt[3]{-571787}=\sqrt[3]{(-83)^3}= -83$$.

Therefore, option $$B$$ is correct.

On prime factorising, we get,

$$ 175616= 8\times8\times8\times7\times7\times7$$

             $$= 8^3 \times 7^3$$.

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.

On prime factorizing, we get,

$$592704 = 2\times2\times2\times2\times2\times2\times3\times3\times3\times7\times7\times7$$

              $$= 2^3\times2^3\times3^3\times7^3 \\=84^3.$$

Thus, cube root of $$592704$$ is:

$$\sqrt[3]{592704}=\sqrt[3]{84^3 } \\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =84$$.

Therefore, option $$C$$ is correct.

On prime factorising, we get,

$$250047$$ $$=\underline{3\times3\times3}\times\underline{3\times3\times3}\times\underline{7\times 7\times 7}$$

              $$= 3^3\times3^3\times7^3=63^3$$.

Then, cube root of $$250047$$ is:

$$\sqrt[3]{250047}=\sqrt[3]{63^3}= 63$$.

Therefore, option $$A$$ is correct.

On prime factorising, we get,

$$226981=\underline{61\times 61\times 61}$$ $$=61^3$$.

Then, $$-226981$$ $$=(-61)^3$$.

Therefore, value of $$\sqrt[3]{-226981}$$ is:

$$\sqrt[3]{-226981}=\sqrt[3]{(-61)^3}= -61$$.

Therefore, option $$B$$ is correct.

On prime factorising, we get,

$$438976$$ $$=\underline{2\times2\times2}\times\underline{2\times2\times2}\times\underline{19\times 19\times 19}$$

              $$= 2^3\times2^3\times19^3=76^3$$.

Then, cube root of $$438976$$ is:

$$\sqrt[3]{438976}=\sqrt[3]{76^3}= 76$$.

Therefore, option $$C$$ is correct.