9th math ex1.2 new book | Notes Punjab
9th math ex1.2 new book notes. Lets check how to solve grade 9 math chapter 1 exercise 1.2. Class 9 math exercise 1.2 from the Punjab board’s new book is a part of chapter 1 and solution of real numbers will be explained here. why we use laws of indices to solve the questions? Lets first check what are radical expressions and then we will check about surds and its types. we will find that how to rationalize the denominator? How to simplify the sums with the help of laws of indices? you will be able to answer the above mentioned question after checking the written and visual solution of 9th math ex1.2.
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Q 1 Rationalize the denominator of the following:
$$\boldsymbol(\boldsymbol i\boldsymbol)\boldsymbol\;\boldsymbol\;\frac{\mathbf{13}}{\mathbf4\boldsymbol+\sqrt{\mathbf3}}\\\\Rationalize\;with\;the\;conjugate\;of\;\\4+\sqrt3\;\;i.e\;\;4-\sqrt3\\\\\frac{13}{4+\sqrt3}\times\frac{4-\sqrt3}{4-\sqrt3}\\\\=\;\frac{3(4-\sqrt3)}{{(4)}^2\;-\;{(\sqrt3)}^2}\\\\=\;\frac{13(4-\sqrt{13})}{16-3}\\\\=\;\frac{13(4-\sqrt{13})}{13}\\\\=\;4-\sqrt{13}$$ $$\boldsymbol(\boldsymbol i\boldsymbol i\boldsymbol)\boldsymbol\;\boldsymbol\;\frac{\sqrt{\mathbf2}\boldsymbol\;\boldsymbol+\sqrt{\mathbf5}}{\sqrt{\mathbf3}}\\Rationalize\;with\;the\;\sqrt3\;\\\\\frac{\sqrt2\;+\sqrt5}{\sqrt3}\times\frac{\sqrt3\;}{\sqrt3}\\\\=\;\frac{(\sqrt2\;+\sqrt5)\;\sqrt3}{\;{(\sqrt3)}^2}\\\\\\=\;\frac{\sqrt6\;+\sqrt{15})}3$$ $$\boldsymbol(\boldsymbol i\boldsymbol i\boldsymbol i\boldsymbol)\boldsymbol\;\boldsymbol\;\frac{\sqrt{\mathbf2}\boldsymbol\;\boldsymbol-\mathbf1}{\sqrt{\mathbf5}}\\Rationalize\;with\;the\;\sqrt5\;\\\\\frac{\sqrt2\;-1}{\sqrt5}\times\frac{\sqrt5\;}{\sqrt5}\\\\=\;\frac{(\sqrt2\;-1)\;\sqrt5}{\;{(\sqrt5)}^2}\\\\\\=\;\frac{\sqrt{10}\;-\sqrt5)}5$$ $$\boldsymbol(\boldsymbol i\boldsymbol v\boldsymbol)\boldsymbol\;\boldsymbol\;\frac{\mathbf6\boldsymbol-\mathbf4\sqrt{\mathbf2}}{\mathbf6\boldsymbol+\mathbf4\sqrt{\mathbf2}}\\\\Rationalize\;with\;the\;conjugate\;of\\6+4\sqrt{2\;\;\;\;}\;i.e\;\;\;\;\;6-4\sqrt2\\\\\;\frac{6-4\sqrt2}{6+4\sqrt2}\times\;\frac{6-4\sqrt2}{6-4\sqrt2}\\\\=\;\;\frac{{(6-4\sqrt2)}^2}{{(6)}^2-{(4\sqrt2)}^2}\;\\\\=\;\frac{{(6)}^2+{(4\sqrt2)}^2-2(6)(4\sqrt2)}{36-(16\times2)}\\\\=\;\frac{36+(16\times2)-48\sqrt2}{36-32}\\\\=\;\frac{36+32-48\sqrt2}4\\\\=\;\frac{68-48\sqrt2}4\\\\=\;\frac{4(17-12\sqrt2)}4\\\\=\;17-12\sqrt2\\$$
$$\boldsymbol(\boldsymbol v\boldsymbol)\boldsymbol\;\boldsymbol\;\frac{\sqrt{\mathbf3}\boldsymbol-\sqrt{\mathbf2}}{\sqrt{\mathbf3}\boldsymbol+\sqrt{\mathbf2}}\\Rationalize\;with\;the\;conjugate\\of\;\sqrt3+\sqrt{2\;\;}\;i.e\;\;\;\sqrt3-\sqrt2\\\\\boldsymbol\;\frac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}\times\;\frac{\sqrt3-\sqrt2}{\sqrt3-\sqrt2}\\\\=\;\;\frac{{(\sqrt3-\sqrt2)}^2}{{(\sqrt3)}^2-{(\sqrt2)}^2}\;\\\\=\;\frac{{(\sqrt3)}^2+{(\sqrt2)}^2-2(\sqrt3)(\sqrt2)}{3-2}\\\\=\;\frac{3+2-2\sqrt6}1\\\\=\;\frac{5-2\sqrt6}1\\\\=\;5-2\sqrt6$$ $$\boldsymbol(\boldsymbol v\boldsymbol i\boldsymbol)\boldsymbol\;\boldsymbol\;\frac{\mathbf4\sqrt{\mathbf3}}{\sqrt{\mathbf7}\boldsymbol+\sqrt{\mathbf5}}\\Rationalize\;with\;the\;conjugate\\of\;\sqrt7+\sqrt{5\;}\;\;i.e\;\;\;\sqrt7-\sqrt5\\\\\boldsymbol\;\frac{4\sqrt3}{\sqrt7+\sqrt5}\times\;\frac{\sqrt7-\sqrt5}{\sqrt7-\sqrt5}\\\\\\=\;\;\frac{4\sqrt3(\sqrt7-\sqrt5)}{{(\sqrt7)}^2-{(\sqrt5)}^2}\;\\\\\\=\;\frac{{4(\sqrt{21}}-\sqrt{15})}{7-5}\\\\=\;\frac{{4(\sqrt{21}}-\sqrt{15})}2\\\\=\;{2(\sqrt{21}}-\sqrt{15})$$
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Q 2 Simplify the following:
$$(i)\;\;\;{(\frac{86}{16})}^\frac{-3}4\\\\=\;\;{(\frac{16}{86})}^\frac34\;\\\\=\;\frac{{(16)}^\frac34}{\;{(86)}^\frac34}\\\\=\;\frac{{\;(2^4)}^{{}^\frac34}}{{(3^4)}^\frac34}\;\\\\=\;\frac{{(2)}^{4\times\frac34}}{{(3)}^{4\times\frac34}}\\\\=\;\frac{{(2)}^3}{{(3)}^3}\\\\=\;\frac8{27}\\$$
$$(ii)\;{\;\;\;\;\;(\frac34)}^{-2}\;\div\;\;{(\frac49)}^3\;\times\;\frac{16}{27}\\\\=\;{\;\;(\frac43)}^2\;\times\;\frac1{\;{(\frac49)}^{3\;}}\times\;\frac{16}{27}\\\\=\;\frac{16}9\times\frac1{\displaystyle\frac{64}{729}}\;\times\;\frac{16}{27}\\\\=\;\frac{16}9\times\frac{729}{\displaystyle64}\;\times\;\frac{16}{27}\\\\=\;12\\$$
$$(iii)\;\;\;{(0.027)}^\frac{-1}3\\\\=\;\;{(\frac{0027}{1000})}^\frac{-1}3\\\\=\;\;{(\frac{1000}{27})}^\frac13\\\\=\;\;\frac{{(10^3)}^{\displaystyle\frac13}}{{(3^3)}^\frac13}\\\\=\;\;\frac{\sqrt[3]{10^3}}{\sqrt[3]{3^3}}\\\\=\;\;\frac{10}3$$
$$(iv)\;\;\;\sqrt[7]{\frac{x^{14}\times y^{21}\times z^{35}}{y^{14}\;z^7}}\\\\=\;\;\sqrt[7]{x^{14}\times y^{21}\times z^{35}\;\times y^{-14}\times z^{-7}}\\\\=\;\;\sqrt[7]{x^{14}\times y^{21-14}\times z^{35-7}\;}\\\\=\;\;\sqrt[7]{x^{14}\times y^7\times z^{28}}\\\\=\;\;\sqrt[7]{x^{14}}\;\times\;\sqrt[7]{y^7}\;\times\;\sqrt[7]{z^{28}}\\\\=\;{(x^{14})}^\frac17\;\times\;\;{(y^7)}^\frac17\times\;\;{(z^{28})}^\frac17\\\\=\;{(x)}^{14\times\frac17}\times\;\;\;{(y)}^{7\times\frac17}\;\times\;\;{(z)}^{28\times\frac17}\\\\=\;x^2\;y\;z^4$$
$$(v)\;\;\;\frac{5.{(25)}^{n+1}-25.{(5)}^{2n}}{5.{(5)}^{2n+3}-{(25)}^{n+1}}\\\\=\;\;\frac{5.{(5^2)}^{n+1}-5^2\;{(5)}^{2n}\;}{5.{(5)}^{2n+3}\;-{(5^2)}^{n+1}}\\\\=\;\;\frac{5.{(5)}^{2(n+1)}-5^{2+2n}\;}{5^{1+2n+3}\;-{(5)}^{2(n+1)}}\\\\=\;\frac{5^{1+2n+2}-5^{2+2n}\;}{5^{4+2n}\;-5^{2n+2}}\\\\=\;\frac{5^{3+2n}-5^{2+2n}\;}{5^{4+2n}\;-5^{2n+2}}\\\\=\;\frac{5^{2n}(5^3-5^2)}{5^{2n}(5^4-5^2)}\\\\=\;\frac{(5^3-5^2)}{(5^4-5^2)}\\\\=\frac{125-25}{625-25}\\\\=\frac{100}{600}\\\\=\frac16\\$$
$$(vi)\;\;\;\frac{{(16)}^{x+1}+20(4^{2x})}{2^{x-3}\times8^{x+2}}\\\\\\=\;\;\frac{{(2^4)}^{x+1}+(2^2\times5^1){(2^2)}^{2x}}{2^{x-3}\times{(2^3)}^{x+2}}\\\\\\=\;\;\frac{{(2)}^{4(x+1)}+(2^2\times5^1){(2)}^{2(2x)}}{2^{x-3}\times{(2)}^{3(x+2)}}\\\\\\=\;\;\frac{2^{4x+4}+2^2\times5{\times2}^{4x}}{2^{x-3}\times2^{3x+6}}\\\\\\=\;\;\frac{2^{4x+4}+2^{2+4x}\times5}{2^{x-3+3x+6}}\\\\\\=\;\;\frac{2^{4x}(2^4+2^2\times5)}{2^{4x+3}}\\\\\\=\;\;\frac{16+4\times5}{2^3}\\\\\\=\;\;\frac{36}8\;=\;\frac92$$
$$(vii)\;\;\;\;\;\;\;{(64)}^\frac{-2}3\div\;\;{(9)}^\frac{-3}2\\\\=\;\;\;{(4^3)}^\frac{-2}3\div\;{(3^2)}^\frac{-3}2\\\\=\;\;4^{3\times\frac{-2}3}\div\;3^{2\times\frac{-3}2}\\\\=\;\;4^{-2}\;\div\;3^{-3}\\\\=\;\;4^{-2}\;\times\;3^3\\\\=\;\frac1{4^2}\times\;3^3\\\\=\;\frac{27}{16}\\$$
$$(viii)\;\;\;\frac{3^n\times\;9^{n+1}}{3^{n-1}\times\;9^{n-1}}\\\\=\;\;\frac{3^n\times\;{(3^2)}^{n+1}}{3^{n-1}\times\;{(3^2)}^{n-1}}\\\\=\;\;\frac{3^n\times\;{(3)}^{2(n+1)}}{3^{n-1}\times\;{(3)}^{2(n-1)}}\\\\=\;\;\;\frac{3^n\times\;{(3)}^{2n+2}}{3^{n-1}\times\;{(3)}^{2n-2}}\\\\=\;\;\frac{3^{n+2n+2}}{3^{n-1+2n-2}}\\\\=\;\;\frac{3^{3n+2}}{3^{3n-3}}\\\\=\;\;3^{3n+2}\;\times\;3^{-3n+3}\\\\=\;\;3^{3n+2-3n+3}\\\\=\;3^5\;=\;243\\$$
$$(ix)\;\;\;\frac{5^{n+3}-6\times5^{n+1}}{9\times5^n-4\times5^n}\\\\=\;\;\;\frac{5^{n+1}(5^2-6)}{5^n(9-4)}\\\\=\;\;\;\frac{5^{n+1}(25-6)}{5^n(5)}\\\\=\;\frac{5^{n+1}(19)}{5^{n+1}}\\\\=\;19$$
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Q 3 $$x\;=\;3\;+\;\sqrt8\;\;\;\\\\Find\;the\;values\;of$$
$$x\;=\;3\;+\;\sqrt8\\(i)\;\;x+\frac1x\;=\;?\\Solution:\\x\;=\;3\;+\;\sqrt8\\\frac1x\;=\;\frac1{3\;+\;\sqrt8}\\\\$$
$$\frac1x\;=\;\frac1{3\;+\;\sqrt8}\times\frac{3\;-\;\sqrt8}{3\;-\;\sqrt8}\\\\\\\frac1x\;=\;\frac{3\;-\;\sqrt8}{{(3)}^2\;-\;{(\sqrt8)}^2}\\\\\frac1x\;=\;\frac{3\;-\;\sqrt8}{9\;-8}\\\\\frac1x=\;3\;-\;\sqrt8\\$$
$$x+\frac1x=\;3\;+\;\sqrt8\;+\;3\;-\;\sqrt8\\x+\frac1x=\;6$$
$$(ii)\;\;x-\frac1x=\;?\\\\\;\;x-\frac1x=\;3\;+\;\sqrt8\;-\;(\;3\;-\;\sqrt8\;)\\\\x-\frac1x=\;\;3\;+\;\sqrt8\;-\;\;3\;+\;\sqrt8\;\\\\x-\frac1x=\;\;2\sqrt8$$
$$(iii)\;\;\;\;x^2\;+\frac1{x^2}\;=\;?$$
$$x+\frac1x\;=\;6\\\\Taking\;square\;on\;both\;the\;sides\\\\\\{(x+\frac1x)}^2\;=\;{(6)}^2\\\\x^2+\frac1{x^2}+2\;=\;36\\\\x^2+\frac1{x^2}=\;36-2\\\\x^2+\frac1{x^2}=\;34\\\\$$
$$(iv)\;\;\;x^2-\;\frac1{x^2}\;=\;?\\We\;have\;already\;found\\\\x+\;\frac1x\;=\;\;6\\x-\frac1x\;=\;2\sqrt8\\\\Multiplying\;the\;above\;two\;equations\\\\(x+\;\frac1x)\;(x-\frac1x)\;=\;(6)\;(2\sqrt8)\\\\x^2\;-\;\frac1{x^2}\;=\;12\sqrt8\\\\$$
$$(v)\;\;\;x^4\;+\;\frac1{x^4}\;=\;?\\\\$$
$$We\;have\;already\;found\\\\x^2\;+\;\frac1{x^2}\;=\;34\\Taking\;square\;on\;both\;the\;side\\\\{(x^2\;+\;\frac1{x^2})}^2\;=\;{(34)}^2\;\;\\x^4\;+\;\frac1{x^4}\;+2\;=\;1156\\\\\\x^4\;+\;\frac1{x^4}\;\;=\;1156-2\\\\x^4\;+\;\frac1{x^4}\;\;=\;1154$$
$$(vi)\;\;\;{(x-\frac1x)}^2\;=\;?$$
$$We\;have\;already\;found\\\\x-\frac1x=2\sqrt8\\\\Taking\;square\;on\;both\;the\;sides\\\\\;{(x-\frac1x)}^2\;=\;{(2\sqrt8)}^2\;\;\;or\;\;{(x-\frac1x)}^2\;=\;4\times8\\\\\;{(x-\frac1x)}^2\;=\;32$$
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Q 4 Find rational numbers p and q such that $$\frac{8-3\sqrt2}{4+3\sqrt2}\;=\;p\;+\;q\sqrt2$$
$$\frac{8-3\sqrt2}{4+3\sqrt2}\;=\;p\;+\;q\sqrt2\\\\Rationalize\;with\;the\;conjugate\;of\;4+3\sqrt2\\i.e\;\;\;\;4-3\sqrt2$$
$$\frac{8-3\sqrt2}{4+3\sqrt2}\;=\;p\;+\;q\sqrt2\\\\\frac{8-3\sqrt2}{4+3\sqrt2}\;\times\;\frac{4-3\sqrt2}{4-3\sqrt2}\;=\;p\;+\;q\sqrt2\\\\\frac{(8-3\sqrt2)\;(4-3\sqrt2)}{(4+3\sqrt2)\;(4-3\sqrt2)}\;=\;\;p\;+\;q\sqrt2\\\\\frac{32-24\sqrt2-12\sqrt2+9\sqrt4}{{(4)}^2\;-\;{(3\sqrt2)}^2}\;=\;p\;+\;q\sqrt2\\\\\frac{32-36\sqrt2+9(2)}{16-\;{(9\times2)}}\;=\;p\;+\;q\sqrt2\\\\\frac{32-36\sqrt2+18}{16-18}\;=\;p\;+\;q\sqrt2\\\\\frac{50-36\sqrt2}{-2}\;=\;p\;+\;q\sqrt2\\\\\frac{-2(-25+18\sqrt2\;)}{-2}\;=\;p\;+\;q\sqrt2\\\\-25+18\sqrt2\;=\;p\;+\;q\sqrt2\\\\Equating\\\\p\;=\;-25\;\;\;and\;\;\;q\;=\;18\\\\\\\\$$
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Q 5 (Important question) Simplify the following:
$$\boldsymbol(\boldsymbol i\boldsymbol)\boldsymbol\;\boldsymbol\;\boldsymbol\;\boldsymbol\;\boldsymbol\;\frac{{\boldsymbol(\mathbf{25}\boldsymbol)}^{\displaystyle\frac{\mathbf3}{\mathbf2}}\boldsymbol\times\boldsymbol\;{\boldsymbol(\mathbf{243}\boldsymbol)}^{\displaystyle\frac{\mathbf3}{\mathbf5}}}{{\boldsymbol(\mathbf{16}\boldsymbol)}^{\displaystyle\frac{\mathbf5}{\mathbf4}}\boldsymbol\times{\boldsymbol(\mathbf8\boldsymbol)}^{\displaystyle\frac{\mathbf4}{\mathbf3}}}\\\\$$
$$=\;\frac{{(25)}^{\displaystyle\frac32}\times\;{(243)}^{\displaystyle\frac35}}{{(16)}^{\displaystyle\frac54}\times{\;(8)}^{\displaystyle\frac43}}\;\\\\=\frac{{(5^2)}^{\displaystyle\frac32}\times\;{(3^5)}^{\displaystyle\frac35}}{{(2^4)}^{\displaystyle\frac54}\times\;{{(2^3})}^{\displaystyle\frac43}}\\\\=\;\frac{{(5)}^{2\times{\displaystyle\frac32}}\times\;{\;{(3)}}^{5\times{\displaystyle\frac35}}}{{(2)}^{4\times{\displaystyle\frac54}}\times\;\;{(2)}^{3\times{\displaystyle\frac43}}}\\\\=\;\frac{5^3\;\times\;3^3}{2^5\;\times\;2^4}\\\;\\=\;\;\frac{125\;\times\;27}{32\;\times\;16}\\\\=\;\frac{3375}{512}$$
$$\boldsymbol(\boldsymbol i\boldsymbol i\boldsymbol)\boldsymbol\;\boldsymbol\;\boldsymbol\;\frac{\mathbf{54}\boldsymbol\;\boldsymbol\times\boldsymbol\;\sqrt[\mathbf3]{{\boldsymbol(\mathbf{27}\boldsymbol)}^{\mathbf2\mathbf x}}}{\mathbf9^{\mathbf x\boldsymbol+\mathbf1}\boldsymbol+\boldsymbol\;\mathbf{216}\boldsymbol\;\boldsymbol(\boldsymbol\;\mathbf3^{\mathbf2\mathbf x\boldsymbol-\mathbf1}\boldsymbol\;\boldsymbol)}$$
$$=\;\;\frac{54\;\times\;\sqrt[3]{{(27)}^{2x}}}{9^{x+1}+\;216\;(\;3^{2x-1}\;)}\;\\\\=\;\;\frac{2\times27\times\sqrt[3]{{(3}^3)^{2x}}}{{{(3}^2)}^{x+1}+8\times27\;(\;3^{2x-1}\;)\;}\\\\=\;\frac{2\times3^3\times\sqrt[3]{{(3)}^{3(2x)}}}{{(3)}^{2(x+1)}+2^3\times3^3(3^{2x-1}\;)\;}\\\\=\;\;\frac{2\times3^3\times{(3^{6x})}^{\displaystyle\frac13}}{{(3)}^{2x+2}+\;2^3\times{(3)}^{3+2x-1}}\\\\=\;\frac{2\times3^3\times{(3)}^{(6x)({\displaystyle\frac13})}}{{(3)}^{2x+2}+\;2^3\times{(3)}^{2x+2}}\\\\\\=\;\frac{2\times3^3\times{(3)}^{2x}}{{(3)}^{2x+2}(\;1+\;2^3\;)}\\\\=\;\frac{2\times{(3)}^{2x+3}}{{(3)}^{2x+2}(\;1+\;8\;)}\\\\=\;\frac{2\times{(3)}^{2x+2}\times3^1}{{(3)}^{2x+2}(\;9\;)}\\\\=\frac{2\times3}9\\\\=\;\frac23\\$$
$$\boldsymbol(\boldsymbol i\boldsymbol i\boldsymbol i\boldsymbol)\boldsymbol\;\boldsymbol\;\boldsymbol\;\boldsymbol\;\sqrt{\boldsymbol\;\frac{\boldsymbol\;{\boldsymbol(\mathbf{216}\boldsymbol)}^\frac{\mathbf2}{\mathbf3}\boldsymbol\times{\boldsymbol(\mathbf{25}\boldsymbol)}^\frac{\mathbf1}{\mathbf2}}{{\boldsymbol(\mathbf0\boldsymbol.\mathbf{04}\boldsymbol)}^\mathbf{\displaystyle\frac{-3}2}}}$$
$$=\;\;\;\sqrt{\;\frac{\;{(216)}^\frac23\times{(25)}^\frac12}{{(0.04)}^{\displaystyle\frac{-3}2}}}\;\\\\=\;\sqrt{\frac{{(6^3)}^\frac23\;\times\;{(5^2)}^\frac12\;}{\;{(\frac{004}{100})}^\frac{-3}2}}\\\\=\;\sqrt{\frac{{(6)}^{3\times{\displaystyle\frac23}}\;\times\;{(5)}^{2\times{\displaystyle\frac12}}\;}{\;{(\frac1{25})}^\frac{-3}2}}\\\\=\;\sqrt{\frac{{(6)}^{3\times{\displaystyle\frac23}}\;\times\;{(5)}^{2\times{\displaystyle\frac12}}\;}{\;{(\frac1{25})}^\frac{-3}2}}\\\\=\;\sqrt{\frac{6^2\;\times\;5}{\;{(25)}^\frac32}}\\\\=\;\sqrt{\frac{6^2\;\times\;5}{\;{(5^2)}^{\displaystyle\frac32}}}\\\\=\;\sqrt{\frac{6^2\;\times\;5}{\;{(5)}^{2\times{\displaystyle\frac32}}}}\\\\=\;\sqrt{\frac{6^2\;\times\;5}{\;5^3}}\\\\=\sqrt{\frac{36\;\times\;5}{\;125}}\\\\=\sqrt{\frac{36}{25}}\\\\=\;\frac65$$
$$\boldsymbol(\boldsymbol i\boldsymbol v\boldsymbol)\boldsymbol\;\boldsymbol(\boldsymbol a^\frac{\mathbf1}{\mathbf3}\boldsymbol+\boldsymbol b^\frac{\mathbf2}{\mathbf3}\boldsymbol)\boldsymbol\times\boldsymbol(\boldsymbol a^\frac{\mathbf2}{\mathbf3}\boldsymbol-\boldsymbol a^\frac{\mathbf1}{\mathbf3}\boldsymbol b^\frac{\mathbf2}{\mathbf3}\boldsymbol+\boldsymbol b^\frac{\mathbf4}{\mathbf3}\boldsymbol)\\$$
$$=(a^\frac13+b^\frac23)\;(a^\frac23-a^\frac13b^\frac23+b^\frac43)\\\\=a^\frac13(\;a^\frac23-a^\frac13b^\frac23+b^\frac34)\\+b^\frac23(a^\frac23-a^\frac13b^\frac23+b^\frac43)\\\\\\=a^\frac13a^\frac23-a^\frac13a^\frac13b^\frac23+a^\frac13b^\frac34\\+b^\frac23a^\frac23-b^\frac23a^\frac13b^\frac23+b^\frac23b^\frac43\\$$
‘$$=\;a^{\frac13+\frac23}-\;a^{\frac13+\frac13}b^\frac23+\;a^\frac13b^\frac34\;\\+\;\;b^\frac23a^\frac23-\;b^{\frac23+\frac23}a^\frac13+b^{\frac23+\frac43}\\\\\\=\;a^1-a^\frac23\;b^\frac23+a^\frac13b^\frac34\;\\+\;\;b^\frac23a^\frac23-\;b^\frac43\;a^\frac13\;+\;b^2\\$$
$$=\;a\;-\;a^\frac23\;b^\frac23\;+\;a^\frac13\;b^\frac43\;\\\;+\;a^\frac23\;b^\frac23\;-\;a^\frac13\;b^\frac43\;+\;b^2\;\\\\\\=a\;+\;b^2\\$$
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