Question

Find the cube of 216, -1331, 1728,
+ Find the cube root of $\sqrt[\tt{3}]{\tt{0.125}}$

Answer 1

Answer:

$\huge\color {pink}\boxed{\colorbox{green} {Aɴsᴡᴇʀシ︎}}$

What is the Cube Root of 1728?

The cube root of 1728 is the number which when multiplied by itself three times gives the product as 1728. Since 1728 can be expressed as 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3. Therefore, the cube root of 1728 = ∛(2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3) = 12.

How to Calculate the Value of the Cube Root of 1728?

Cube Root of 1728 by Prime Factorization

Prime factorization of 1728 is 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3

Simplifying the above expression: 26 × 33

Simplifying further: 123

Therefore, the cube root of 1728 by prime factorization is (2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3)1/3 = 12.

Answer 2

To Find: The cube of 216

We have to find the factors of 216.

          $\Large{ \begin{array}{c|c} \tt 2 & \sf{ 216} \\ \tt 2 & \sf { 108} \\ \tt 2 & \sf{ 54} \\ \tt 3 & \sf{ 27} \\ \tt 3 & \sf{ 9 }\\ \tt 3 & \sf{ 3 }\\ \tt & \sf{ 1} \end{array}}$

216,

$\implies \sf{2 \times 2 \times 2 \times 3 \times 3 \times 3 }$

$\implies \sf{(2 \times 3) \times (2 \times 3) \times (2 \times 3)}$

$\implies \sf{(2 \times 3)^3}$

$\implies \sf{(6)^3}$

$\therefore \underline{\boxed{ \purple {\sf{\sqrt[\sf{3}]{\sf{216 \; }} = 6 }}}}$

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To Find: The cube of -1331

We have to find the factors of -1331

          $\Large{ \begin{array}{c|c} \tt 11 & \sf{ 1331} \\ \tt 11 & \sf { 121} \\ \tt 11 & \sf{ 11} \\ \tt & \sf{ 1} \end{array}}$

1331,

$\implies \sf{11 \times 11 \times 11}$

$\implies \sf{(11)^3}$

$\therefore \underline{\boxed{ \purple {\sf{\sqrt[\sf{3}]{\sf{-1331 \; }} = -11 }}}}$

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To Find: The cube of 1728

We have to find the factors of 1728

          $\Large{ \begin{array}{c|c} \tt 2 & \sf{ 1728} \\ \tt 2 & \sf { 864} \\ \tt 2 & \sf{ 432} \\ \tt 2 & \sf{ 216} \\ \tt 2 & \sf{ 108 }\\ \tt 2 & \sf{ 54 }\\ \tt 3 & \sf{ 27} \\ \tt 3 & \sf{ 9 }\\ \tt 3 & \sf{ 3} \\ & \sf{ 1} \end{array}}$

1728,

$\implies \sf{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 }$

$\implies \sf{( 2 \times 3) \times ( 2 \times 3) \times ( 2 \times 3) \times 2^3}$

$\implies \sf{( 2 \times 3)^3 \times 2^3}$

$\implies \sf{(6)^3 \times 2^3}$

$\implies \sf{(6 \times 2)^3} \dots \Bigg \langle \frak{\orange{ a^m \times b^m = (a \times b)^m }} \Bigg \rangle$

$\implies \sf{(12)^3}$

$\therefore \underline{\boxed{ \purple {\sf{\sqrt[\sf{3}]{\sf{1728 \; }} = 12 }}}}$

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To Find: The cube root of ∛0.125

$\implies \sf{\sqrt[\sf{3}]{\sf{ \dfrac{ 125 }{ 1000 } }} }$

$\implies \sf{ \dfrac{\sqrt[3]{\sf{125 \;}} }{ \sqrt[3]{\sf{1000} \;} }} \dots \Bigg \langle \frak{\orange{ \bigg( \dfrac{a}{b} \bigg)^m = \dfrac{a^m}{b^m}}} \Bigg \rangle$

$\implies \sf{ \dfrac{ \sqrt[\sf{3}]{\sf{5^3 \; }} }{ \sqrt[\sf{3}]{\sf{10^3 \; }} } }$

$\implies \sf{\dfrac{5}{10} } \dots \Bigg \langle \frak{\orange{ \bigg[(a^{m})\bigg]^{\dfrac{1}{m}} =a }} \Bigg \rangle$

$\implies \sf{0.5}$

$\therefore \underline{\boxed{ \purple {\sf{\sqrt[\sf{3}]{\sf{0.125 \; }} = 0.5 }}}}$

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Apologies for the mistakes!