# Find the cube roots of each of the following integers:(i) $-125$(ii) $-5832$(iii) $-2744000$(iv) $-753571$(v) $-32768$

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To find:

We have to find the cube roots of the given integers.

Solution:

(i) $-125=-(5\times5\times5)$

$=-(5^3)$

Therefore,

$\sqrt[3]{-125}=\sqrt[3]{-(5)^3}$

$=-5$

(ii) $-5832=-(2\times2\times2\times3\times3\times3\times3\times3\times3)$

$=-[(2^3)\times(3^3)\times(3^3)]$

$=-(2\times3\times3)^3$

$=-(18)^3$

Therefore,

$\sqrt[3]{-5832}=\sqrt[3]{-(18)^3}$

$=-18$

(iii) $-2744000=-(2\times2\times2\times2\times2\times2\times5\times5\times5\times7\times7\times7)$

$=-[(2^3)\times(2^3)\times(5^3)\times(7^3)]$

$=-(2\times2\times5\times7)^3$

$=-(140)^3$

Therefore,

$\sqrt[3]{-2744000}=\sqrt[3]{-(140)^3}$

$=-140$

(iv) $-753571=-(7\times7\times7\times13\times13\times13)$

$=-[(7^3)\times(13^3)]$

$=-(7\times13)^3$

$=-(91)^3$

Therefore,

$\sqrt[3]{-753571}=\sqrt[3]{-(91)^3}$

$=-91$

(v) $-32768=-(2\times2\times2\times2\times2\times2\times2\times2\times2\times2\times2\times2\times2\times2\times2)$

$=-[(2^3)\times(2^3)\times(2^3)\times(2^3)\times(2^3)]$

$=-(2\times2\times2\times2\times)^3$

$=-(32)^3$

Therefore,

$\sqrt[3]{-32768}=\sqrt[3]{-(32)^3}$

$=-32$

Updated on 10-Oct-2022 12:47:19