3.2.2 Squares, Square Roots, Cube and Cube Roots, PT3 Focus Practice - Mathematics

3.2.2 Squares, Square Roots, Cube and Cube Roots, PT3 Focus Practice

Question 5:

Complete the operation steps below by filling in the boxes using suitable numbers.

\begin{array}{l} \sqrt[3]{4 \frac{17}{27}} \div \sqrt{1 \frac{9}{16}} = \sqrt[3]{\frac{}{27}} \div \sqrt{\frac{25}{16}} \\ = \frac{}{3} \div \frac{5}{4} \\ = \frac{}{3} \times \frac{4}{5} \\ = \end{array}

$\begin{array}{l} \sqrt[3]{4 \frac{17}{27}} \div \sqrt{1 \frac{9}{16}} = \sqrt[3]{\frac{}{27}} \div \sqrt{\frac{25}{16}} \\ = \frac{}{3} \div \frac{5}{4} \\ = \frac{}{3} \times \frac{4}{5} \\ = \end{array}$

Solution:

\begin{array}{l} \sqrt[3]{4 \frac{17}{27}} \div \sqrt{1 \frac{9}{16}} = \sqrt[3]{\frac{125}{27}} \div \sqrt{\frac{25}{16}} \\ = \frac{5}{3} \div \frac{5}{4} \\ = \frac{5}{3} \times \frac{4}{5} \\ = \frac{4}{3} \end{array}

$\begin{array}{l} \sqrt[3]{4 \frac{17}{27}} \div \sqrt{1 \frac{9}{16}} = \sqrt[3]{\frac{125}{27}} \div \sqrt{\frac{25}{16}} \\ = \frac{5}{3} \div \frac{5}{4} \\ = \frac{5}{3} \times \frac{4}{5} \\ = \frac{4}{3} \end{array}$

Question 6:

Complete the operation steps below by filling in the boxes using suitable numbers.

\begin{array}{l} {(\sqrt{2 \frac{7}{9}} - \sqrt[3]{\frac{27}{64}})}^{2} = {(\sqrt{\frac{}{9}} - \frac{3}{})}^{2} \\ = {(\frac{}{12})}^{2} \\ = \end{array}

$\begin{array}{l} {(\sqrt{2 \frac{7}{9}} - \sqrt[3]{\frac{27}{64}})}^{2} = {(\sqrt{\frac{}{9}} - \frac{3}{})}^{2} \\ = {(\frac{}{12})}^{2} \\ = \end{array}$

Solution:

\begin{array}{l} {(\sqrt{2 \frac{7}{9}} - \sqrt[3]{\frac{27}{64}})}^{2} = {(\sqrt{\frac{25}{9}} - \frac{3}{4})}^{2} \\ = {(\frac{5}{3} - \frac{3}{4})}^{2} \\ = {(\frac{(5 \times 4) - (3 \times 3)}{12})}^{2} \\ = {(\frac{11}{12})}^{2} \\ = \frac{121}{144} \end{array}

$\begin{array}{l} {(\sqrt{2 \frac{7}{9}} - \sqrt[3]{\frac{27}{64}})}^{2} = {(\sqrt{\frac{25}{9}} - \frac{3}{4})}^{2} \\ = {(\frac{5}{3} - \frac{3}{4})}^{2} \\ = {(\frac{(5 \times 4) - (3 \times 3)}{12})}^{2} \\ = {(\frac{11}{12})}^{2} \\ = \frac{121}{144} \end{array}$

Question 7:

Complete the operation steps below by filling in the boxes using suitable numbers.

\begin{array}{l} \sqrt[3]{1 \frac{61}{64}} - {0.3}^{2} = \sqrt[3]{\frac{}{64}} - {0.3}^{2} \\ = \frac{}{4} - \\ = \end{array}

$\begin{array}{l} \sqrt[3]{1 \frac{61}{64}} - {0.3}^{2} = \sqrt[3]{\frac{}{64}} - {0.3}^{2} \\ = \frac{}{4} - \\ = \end{array}$

Solution:

\begin{array}{l} \sqrt[3]{1 \frac{61}{64}} - {0.3}^{2} = \sqrt[3]{\frac{125}{64}} - {0.3}^{2} \\ = \frac{5}{4} - 0.09 \\ = 1.25 - 0.09 \\ = 1.16 \end{array}

$\begin{array}{l} \sqrt[3]{1 \frac{61}{64}} - {0.3}^{2} = \sqrt[3]{\frac{125}{64}} - {0.3}^{2} \\ = \frac{5}{4} - 0.09 \\ = 1.25 - 0.09 \\ = 1.16 \end{array}$

Question 8:

Find the value of:

\begin{array}{l} (a) \sqrt[3]{\frac{10}{27} - 5} \\ (b) \sqrt{\frac{4}{49}} \times \sqrt[3]{- 0.216} \end{array}

$\begin{array}{l} (a) \sqrt[3]{\frac{10}{27} - 5} \\ (b) \sqrt{\frac{4}{49}} \times \sqrt[3]{- 0.216} \end{array}$

Solution:

\begin{array}{l} (a) \sqrt[3]{\frac{10}{27} - 5} = \sqrt[3]{\frac{10 - 135}{27}} \\ = \sqrt[3]{- \frac{125}{27}} \\ = - \frac{5}{3} \end{array}

$\begin{array}{l} (a) \sqrt[3]{\frac{10}{27} - 5} = \sqrt[3]{\frac{10 - 135}{27}} \\ = \sqrt[3]{- \frac{125}{27}} \\ = - \frac{5}{3} \end{array}$

\begin{array}{l} (b) \sqrt{\frac{4}{49}} \times \sqrt[3]{- 0.216} = \frac{2}{7} \times \sqrt[3]{\frac{216}{1000}} \\ = \frac{2}{7} \times \frac{6}{510} \\ = \frac{6}{35} \end{array}

$\begin{array}{l} (b) \sqrt{\frac{4}{49}} \times \sqrt[3]{- 0.216} = \frac{2}{7} \times \sqrt[3]{\frac{216}{1000}} \\ = \frac{2}{7} \times \frac{6}{510} \\ = \frac{6}{35} \end{array}$

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