Question

Find the cube root of each of the following numbers by prime Factorization method:
(i) $729$ (ii) $343$ (iii) $512$ (iv) $0.064$ (v) $0.216$ (vi) $5\frac{23}{64}$ (vii) $-1.331$ (viii) $-27000$

Answer 1

(i) $729=3^6$

$\therefore \sqrt[3]{3729}=\sqrt[3]{33^6}=3^2=9$

(ii) $343=7^3$

$\therefore \sqrt[3]{3343}=\sqrt[3]{37^3}=7$

(iii) $512=8^3$

$\therefore \sqrt[3]{3512}=\sqrt[3]{38^3}=8$

(iv) $0.064=2^3\times (0.001{)}^3$

$\therefore \sqrt[3]{30.064}=\sqrt[3]{32^3\times (0.1{)}^3}=0.2$

(v) $0.216=2^3\times 3^3\times {0.1}^3$

$\therefore \sqrt[3]{30.216}=\sqrt[3]{32^3\times 3^3\times {0.1}^3}=0.6$

(vi) $5\frac{23}{64}=\frac{343}{64}=\frac{7^3}{2^6}$

$\therefore \sqrt[3]{3\frac{343}{64}}=\sqrt[3]{3\frac{7^3}{2^6}}=\frac{7}{8}$

(vii) $-1.331=-1^3\times {11}^3$

$\therefore \sqrt[3]{3-1331}=\sqrt[3]{3-1^3\times {11}^3}=-11$

(viii) $-27000=-1^3\times 3^3\times 2^3\times 5^3$

$\therefore \sqrt[3]{3-27000}=\sqrt[3]{3-1^3\times 3^3\times 2^3\times 5^3}=-30$