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Question: Find the cube root of 438976 using prime factorization. (1) 74 (2) 72 (3) 76 (4) 71...

Find the cube root of 438976 using prime factorization.
(1) 74
(2) 72
(3) 76
(4) 71

Explanation

Solution

Hint:Divide the number using prime numbers according to divisibility.The method of prime factorization is used to “break down” or express a given number as a product of prime numbers.The prime numbers will occur more than once in the prime factorization and they can be expressed together in an exponential form to look compact where prime number is a whole number which is greater than 1, it is only divisible by 1 and itself.
i.e. prime numbers have exactly two factors namely 1 and itself.

Complete step-by-step answer:
We have the number 438976.

& 2\left| \\!{\underline {\, 438976 \,}} \right. \\\ & 2\left| \\!{\underline {\, 219488 \,}} \right. \\\ & 2\left| \\!{\underline {\, 109744 \,}} \right. \\\ & 2\left| \\!{\underline {\, 54872 \,}} \right. \\\ & 2\left| \\!{\underline {\, 27436 \,}} \right. \\\ & 2\left| \\!{\underline {\, 13718 \,}} \right. \\\ & 19\left| \\!{\underline {\, 6859 \,}} \right. \\\ & 19\left| \\!{\underline {\, 361 \,}} \right. \\\ & 19\left| \\!{\underline {\, 19 \,}} \right. \\\ & \left| \\!{\underline {\, 1 \,}} \right. \\\ \end{aligned}$$ Divide 438976 by prime number 2 until we get 6859. Then number 6859 is divisible by 19 and thus we get 1. $$\begin{aligned} & \therefore 438976=2\times 2\times 2\times 2\times 2\times 2\times 19\times 19\times 19 \\\ & 438976={{2}^{3}}\times {{2}^{3}}\times {{19}^{3}} \\\ \end{aligned}$$ Now, by taking the cube root on both RHS and LHS. $$\begin{aligned} & \sqrt[3]{438976}=\sqrt{{{2}^{3}}\times {{2}^{3}}\times {{19}^{3}}} \\\ & \sqrt[3]{438976}=2\times 2\times 19=76 \\\ \end{aligned}$$ $$\therefore $$The cube root of 438976 = 76 by prime factorization. Hence, the correct option is (c) 76. Note: Another way is to find the cubes of the options provided. a) $74^3 = 405224$ b) $72^3 = 373248$ c) $76^3 = 438976$ d) $71^3 = 357911$