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Find the cube roots of each of the following integers:
(i) $-125$
(ii) $-5832$
(iii) $-2744000$
(iv) $-753571$
(v) $-32768$
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,
$\sqrt[3]{-32768}=\sqrt[3]{-(32)^3}$
$=-32$