Exercise 3.2 — Cube Roots (Full Solutions)
Q1. Find the cube roots of the following numbers
Method: express each number as a product of prime factors in groups of three, then take one factor from each group.
| No. | Number | Prime Factorisation | Cube Root |
|---|---|---|---|
| (a) | 64 | \(2^6 = (2^2)^3\) | \(\sqrt[3]{64}=4\) |
| (b) | 343 | \(7^3\) | \(\sqrt[3]{343}=7\) |
| (c) | 512 | \(2^9=(2^3)^3\) | \(\sqrt[3]{512}=8\) |
| (d) | 2744 | \(2^3\times 7^3\) | \(\sqrt[3]{2744}=14\) |
| (e) | 10648 | \(2^3\times 11^3\) | \(\sqrt[3]{10648}=22\) |
| (f) | 17576 | \(2^3\times 13^3\) | \(\sqrt[3]{17576}=26\) |
| (g) | −1728 | \(-(2^6\times 3^3)=-(12^3)\) | \(\sqrt[3]{-1728}=-12\) |
| (h) | 27000 | \(2^3\times 3^3\times 5^3\) | \(\sqrt[3]{27000}=30\) |
| (i) | 91125 | \(3^6\times 5^3=(45)^3\) | \(\sqrt[3]{91125}=45\) |
| (j) | −35937 | \(-(3^3\times 11^3)=-(33^3)\) | \(\sqrt[3]{-35937}=-33\) |
| (k) | −17576 | \(-(2^3\times 13^3)=-(26^3)\) | \(\sqrt[3]{-17576}=-26\) |
| (l) | \(\dfrac{27}{-4096}\) | \(27=3^3,\ 4096=2^{12}=(16)^3\) | \(\sqrt[3]{\dfrac{27}{-4096}}=\dfrac{3}{-16}=-\dfrac{3}{16}\) |
Q2. Show that the following are true
(a) \(\sqrt[3]{8}\times\sqrt[3]{729}=\sqrt[3]{8\times 729}\)
LHS: \(\sqrt[3]{8}=2,\ \sqrt[3]{729}=9 \Rightarrow 2\times 9=18\)
RHS: \(8\times 729=5832,\ \sqrt[3]{5832}=18\)
(b) \(\sqrt[3]{125}\times\sqrt[3]{343}=\sqrt[3]{125\times 343}\)
LHS: \(\sqrt[3]{125}=5,\ \sqrt[3]{343}=7 \Rightarrow 5\times 7=35\)
RHS: \(125\times 343=42875,\ \sqrt[3]{42875}=35\)
(c) \(\dfrac{\sqrt[3]{729}}{\sqrt[3]{1000}}=\sqrt[3]{\dfrac{729}{1000}}\)
LHS: \(\dfrac{9}{10}\)
RHS: \(\sqrt[3]{0.729}=0.9=\dfrac{9}{10}\)
(d) \(\dfrac{\sqrt[3]{-216}}{\sqrt[3]{1331}}=\sqrt[3]{\dfrac{-216}{1331}}\)
LHS: \(\dfrac{-6}{11}\)
RHS: \(\sqrt[3]{\dfrac{-216}{1331}}=-\dfrac{6}{11}\)
Q3. Evaluate each of the following
(a) \(\sqrt[3]{4^3\times 6^3}=4\times 6=24\)
(b) \(\sqrt[3]{8^3\times 17^3}=8\times 17=136\)
(c) \(\sqrt[3]{700\times 2\times 49\times 5}\)
\(=\sqrt[3]{7^3\times 10^3}\) (since \(700\times 2\times 49\times 5 = 343000 = 7^3\times 10^3\))
\(=7\times 10=70\)
(d) \(\sqrt[3]{27}+\sqrt[3]{0.008}+\sqrt[3]{0.064}\)
\(=3+0.2+0.4=3.6\)
(e) \(\sqrt[3]{1000}+\sqrt[3]{0.008}-\sqrt[3]{0.125}\)
\(=10+0.2-0.5=9.7\)
(f) \(\sqrt[3]{\dfrac{729}{216}}\times \dfrac{6}{9}\)
\(\sqrt[3]{\dfrac{729}{216}}=\dfrac{9}{6}=\dfrac{3}{2}\)
\(=\dfrac{3}{2}\times\dfrac{6}{9}=\dfrac{3}{2}\times\dfrac{2}{3}=1\)
Q4. Find the cube roots using the given factorisations
(a) \(2460375=3375\times 729=15^3\times 9^3=(15\times 9)^3=135^3\)
(b) \(20346417=9261\times 2197=21^3\times 13^3=(21\times 13)^3=273^3\)
(c) \(210644875=42875\times 4913=35^3\times 17^3=(35\times 17)^3=595^3\)
(d) \(57066625=166375\times 343=55^3\times 7^3=(55\times 7)^3=385^3\)
Q5. Volume of a cubical box = 13.824 m³. Find the side length
Side \(=\sqrt[3]{13.824}=\sqrt[3]{\dfrac{13824}{1000}}\)
\(13824=2^9\times 3^3=(24)^3\), and \(1000=10^3\)
Q6. Divide 26244 by the smallest number so the quotient is a perfect cube
Prime factorisation: \(26244=2^2\times 3^8\)
Group in threes: \(2^2\) is one short of a triple (extra pair), \(3^8=3^6\times 3^2\) leaves an extra \(3^2\).
To make the quotient a perfect cube, we must remove the un-grouped factors \(2^2\times 3^2=4\times 9=36\).
Quotient = \(26244\div 36=729=3^6=9^3\)
Cube root of quotient = \(\sqrt[3]{729}=9\)
Q7. Smallest number to multiply 137592 by, to make a perfect cube
Prime factorisation: \(137592=2^3\times 3^3\times 7^2\times 13^1\)
\(2^3\) and \(3^3\) are already complete triples. \(7^2\) needs one more 7; \(13^1\) needs two more 13’s.
Smallest multiplier \(=7\times 13\times 13=7\times 169=1183\)
Product = \(137592\times 1183=162771336=2^3\times 3^3\times 7^3\times 13^3=(2\times 3\times 7\times 13)^3=546^3\)
Cube root of product = \(\sqrt[3]{162771336}=546\)