3.2.3 Squares, Square Roots, Cube and Cube Roots, PT3 Practice - Mathematics

Question 9:

Find the values of the following:

\begin{matrix} (a) {(9 + \sqrt[3]{- 8})}^{2} (b) \sqrt{144} \div \sqrt[3]{216} \times {0.3}^{3} \end{matrix}

$\begin{array}{l} (a) {(9 + \sqrt[3]{- 8})}^{2} \\ (b) \sqrt{144} \div \sqrt[3]{216} \times {0.3}^{3} \end{array}$

Solution:

\begin{matrix} (a) {(9 + \sqrt[3]{- 8})}^{2} = {[9 + (- 2)]}^{2} = 7^{2} = 49 \end{matrix}

$\begin{array}{l} (a) {(9 + \sqrt[3]{- 8})}^{2} = {[9 + (- 2)]}^{2} \\ = 7^{2} \\ = 49 \end{array}$

\begin{matrix} (b) \sqrt{144} \div \sqrt[3]{216} \times {0.3}^{3} = 12 \div 6 \times (0.3 \times 0.3 \times 0.3) = 2 \times 0.027 = 0.054 \end{matrix}

$\begin{array}{l} (b) \sqrt{144} \div \sqrt[3]{216} \times {0.3}^{3} \\ = 12 \div 6 \times (0.3 \times 0.3 \times 0.3) \\ = 2 \times 0.027 \\ = 0.054 \end{array}$

Question 10:

\begin{matrix} (a) Find the value of \sqrt{0.09} (b) Calculate the value of {(\frac{5}{7} \times \sqrt{49})}^{2} \end{matrix}

$\begin{array}{l} (a) Find the value of \sqrt{0.09} \\ (b) Calculate the value of {(\frac{5}{7} \times \sqrt{49})}^{2} \end{array}$

Solution:

\begin{matrix} (a) \sqrt{0.09} = \sqrt{\frac{9}{100}} = \frac{3}{10} = 0.3 (b) {(\frac{5}{7} \times \sqrt{49})}^{2} = {(\frac{5}{7} \times 7)}^{2} = 5^{2} = 25 \end{matrix}

$\begin{array}{l} (a) \\ \sqrt{0.09} = \sqrt{\frac{9}{100}} = \frac{3}{10} = 0.3 \\ (b) \\ {(\frac{5}{7} \times \sqrt{49})}^{2} = {(\frac{5}{7} \times 7)}^{2} \\ = 5^{2} \\ = 25 \end{array}$

Question 11:

\begin{matrix} (a) Find the value of \sqrt[3]{- \frac{1}{27}} (b) Calculate the value of {(7 + \sqrt[3]{- 8})}^{2} \end{matrix}

$\begin{array}{l} (a) Find the value of \sqrt[3]{- \frac{1}{27}} \\ (b) Calculate the value of {(7 + \sqrt[3]{- 8})}^{2} \end{array}$

Solution:

\begin{matrix} (a) \sqrt[3]{- \frac{1}{27}} = \sqrt[3]{- \frac{1}{3} \times - \frac{1}{3} \times - \frac{1}{3}} = - \frac{1}{3} (b) {(7 + \sqrt[3]{- 8})}^{2} = {[7 + (- 2)]}^{2} = 5^{2} = 25 \end{matrix}

$\begin{array}{l} (a) \\ \sqrt[3]{- \frac{1}{27}} = \sqrt[3]{- \frac{1}{3} \times - \frac{1}{3} \times - \frac{1}{3}} = - \frac{1}{3} \\ (b) \\ {(7 + \sqrt[3]{- 8})}^{2} = {[7 + (- 2)]}^{2} \\ = 5^{2} \\ = 25 \end{array}$

Question 12:

\begin{matrix} (a) Find the value of {(- \frac{1}{4})}^{3} . (b) Calculate the value of {(4.2 \div \sqrt[3]{27})}^{2} . \end{matrix}

$\begin{array}{l} (a) Find the value of {(- \frac{1}{4})}^{3} . \\ (b) Calculate the value of {(4.2 \div \sqrt[3]{27})}^{2} . \end{array}$

Solution:

\begin{matrix} (a) {(- \frac{1}{4})}^{3} = - \frac{1}{4} \times - \frac{1}{4} \times - \frac{1}{4} = - \frac{1}{64} (b) {(4.2 \div \sqrt[3]{27})}^{2} = {(4.2 \div 3)}^{2} = {1.4}^{2} = 1.4 \times 1.4 = 1.96 \end{matrix}

$\begin{array}{l} (a) \\ {(- \frac{1}{4})}^{3} = - \frac{1}{4} \times - \frac{1}{4} \times - \frac{1}{4} \\ = - \frac{1}{64} \\ (b) \\ {(4.2 \div \sqrt[3]{27})}^{2} = {(4.2 \div 3)}^{2} \\ = {1.4}^{2} \\ = 1.4 \times 1.4 \\ = 1.96 \end{array}$

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