Question

find the cube root 1404928 with steps

Answer 1

Step-by-step explanation:

1404928 is said to be a perfect cube because 112 x 112 x 112 is equal to 1404928. Since 1404928 is a whole number, it is a perfect cube. The nearest previous perfect cube is 1367631 and the nearest next perfect cube is 1442897

Answer 2

Here, we have to find the cube root of 1404928 , so at first we'll resolve the number into prime factors and then we'll make triplets in order to calculate the cube root of 1404928.

Resolving 1404928 into prime factors we get :

$\large{ \begin{array}{c| c} 2& \underline{1404928} \\ 2& \underline{702464} \\ 2& \underline{351232} \\ 2& \underline{175616} \\ 2& \underline{87808} \\2& \underline{43904} \\2& \underline{21952} \\ 2& \underline{10976} \\ 2& \underline{5488} \\ 2& \underline{2744} \\ 2& \underline{1372} \\ 2& \underline{686} \\ 7& \underline{343} \\ 7& \underline{49} \\ 7& \underline{7} \\ \: &1\end{array}}$

→ 1404928 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 7 × 7 × 7

→ 1404928 = $\sf { {2}^{3} }$ × $\sf { {2}^{3} }$ × $\sf { {2}^{3} }$ × $\sf { {2}^{3} }$ × $\sf { {7}^{3} }$

→ $\small\sqrt[3]{1404928} = \sqrt[3]{ {2}^{3} \times {2}^{3} \times {2}^{3} \times {2}^{3} \times {7}^{3} }$

→ $\small\sqrt[3]{1404928} = \sqrt[3]{ {2}^{3}} \times \sqrt[3]{{2}^{3}} \times \sqrt[3]{{2}^{3} } \times \sqrt[3]{{2}^{3}} \times \sqrt[3]{{7}^{3}}$

[ Since $\sf \purple { \sqrt[3]{a \times b} = \sqrt[3]{a} \times \sqrt[3]{b} }$ ]

→ $\sqrt[3]{1404928}$ = 2 × 2 × 2 × 2 × 7

→ $\underline{\boxed{\sf{\large\sqrt[3]{1404928} = 112 }}} \: \red{\bigstar}$

Hence, cube root of 1404928 is 112.

More :

• The cube of even numbers are always even.
• The cube of odd numbers are always odd.

• ${\underline {\boxed {\Large {\bf \gray {\sqrt[3]{a \times b} = \sqrt[3]{a} \times \sqrt[3]{b} } }}}}$

• ${\underline {\boxed {\Large {\bf \gray {\sqrt[3]{ \dfrac{a }{b}} =\dfrac{ \sqrt[3]{a}}{ \sqrt[3]{b} }} }}}}$