Question

Find the cube root of each of the following numbers by prime factorisation method.

(i) $64$

(ii) $512$

(iii) $10648$

(iv) $27000$

(v) $15625$

(vi) $13824$

(vii) $110592$

(viii) $46656$

(ix) $175616$

(x) $91125$

Answer 1

(i) Prime factorisation of 64=2×2×2×2××2×264=2\times2\times2\times2\times\times2\times2 $64=2\times 2\times 2\times 2\times 2\times 2$

$\sqrt[3]{364}=\sqrt[3]{32^3\times 2^3}=2\times 2=4$

Hence, $\sqrt[3]{364}=4$

(ii) Prime factorisation of $512=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2$

$\sqrt[3]{3512}=\sqrt[3]{32^3\times 2^3\times 2^3}=2\times 2\times 2=8$

Hence, $\sqrt[3]{3512}=8$

(iii) Prime factorisation of $10648=2\times 2\times 2\times 2\times 11\times 11\times 11$

$\sqrt[3]{310648}=\sqrt[3]{32^3\times {11}^3}=2\times 11=22$

Hence, $\sqrt[3]{310648}=22$

(iv) Prime factorisation of $27000=2\times 2\times 2\times 3\times 3\times 3\times 5\times 5\times 5$

$\sqrt[3]{327000}=\sqrt[3]{32^3\times 3^3\times 5^3}=2\times 3\times 5=30$

Hence, $\sqrt[3]{327000}=30$

(v) Prime factorisation of $15625=5\times 5\times 5\times 5\times 5\times 5$

$\sqrt[3]{315625}=\sqrt[3]{35^3\times 5^3}=5\times 5=25$

Hence, $\sqrt[3]{315625}=25$

(vi) Prime factorisation of $13824=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 3\times 3\times 3$

$\sqrt[3]{313824}=\sqrt[3]{32^3\times 2^3\times 2^3\times 3^3}=2\times 2\times 2\times 3=24$

Hence, $\sqrt[3]{313824}=24$

(vii) Prime factorisation of $110592=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 3\times 3\times 3$

$\sqrt[3]{3110592}=\sqrt[3]{32^3\times 2^3\times 2^3\times 3^3}=2\times 2\times 2\times 2\times 3=48$

Hence, $\sqrt[3]{3110592}=48$

(viii) Prime factorisation of $46656=2\times 2\times 2\times 2\times 2\times 2\times 3\times 3\times 3\times 3\times 3\times 3$

$\sqrt[3]{346656}=\sqrt[3]{32^3\times 2^3\times 3^3\times 3^3}=2\times 2\times 3\times 3=36$

Hence, $\sqrt[3]{346656}=36$

(ix) Prime factorisation of $175616=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 7\times 7\times 7$

$\sqrt[3]{3175616}=\sqrt[3]{32^3\times 2^3\times 7^3\times 2^3}=2\times 2\times 2\times 7=56$

Hence, $\sqrt[3]{346656}=56$

(x) Prime factorisation of $91125=3\times 3\times 3\times 3\times 3\times 3\times 5\times 5\times 5$

$\sqrt[3]{391125}=\sqrt[3]{33^3\times 3^3\times 5^3}=3\times 3\times 5=45$

Hence, $\sqrt[3]{391125}=45$