{"id":125014,"date":"2023-09-03T00:55:00","date_gmt":"2023-09-02T19:25:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=125014"},"modified":"2023-10-31T10:37:49","modified_gmt":"2023-10-31T05:07:49","slug":"rd-sharma-class-9-solutions-chapter-4-exercise-4-3","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-4-exercise-4-3\/","title":{"rendered":"RD Sharma Class 9 Solutions Chapter 4 Exercise 4.3 (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone wp-image-125230 size-full\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-4-Exercise-4.3.jpg\" alt=\"RD Sharma Class 9 Solutions Chapter 4 Exercise 4.3\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-4-Exercise-4.3.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-4-Exercise-4.3-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 9 Solutions Chapter 4 Exercise 4.3:<\/strong> Clear your exams with the <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-for-maths\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 9 Maths<\/a>. You can easily download the PDF of <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-4-algebraic-identities\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Solutions Chapter 4<\/a> Exercise 4.3. All the solutions are as per the current CBSE syllabus and are easy to understand. To know more, read the whole blog.\u00a0<\/p>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_47_1 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"ez-toc-toggle-icon-1\"><label for=\"item-69e9287bbf272\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;direction:ltr;\"><svg style=\"fill: #000000;color:#000000\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #000000;color:#000000\" 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class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-4-exercise-4-3\/#access-rd-sharma-class-9-solutions-chapter-4-exercise-43\" title=\"Access RD Sharma Class 9 Solutions Chapter 4 Exercise 4.3\">Access RD Sharma Class 9 Solutions Chapter 4 Exercise 4.3<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-4-exercise-4-3\/#faqs-on-rd-sharma-class-9-solutions-chapter-4-exercise-43\" title=\"FAQs on RD Sharma Class 9 Solutions Chapter 4 Exercise 4.3\">FAQs on RD Sharma Class 9 Solutions Chapter 4 Exercise 4.3<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-4-exercise-4-3\/#how-much-does-it-cost-to-download-the-pdf-of-rd-sharma-solutions-for-class-9-maths-chapter-4-exercise-43\" title=\"How much does it cost to download the PDF of RD Sharma Solutions for Class 9 Maths Chapter 4 Exercise 4.3?\">How much does it cost to download the PDF of RD Sharma Solutions for Class 9 Maths Chapter 4 Exercise 4.3?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-4-exercise-4-3\/#how-many-questions-are-there-in-rd-sharma-solutions-for-class-9-maths-chapter-4-exercise-43\" title=\"How many questions are there in RD Sharma Solutions for Class 9 Maths Chapter 4 Exercise 4.3?\">How many questions are there in RD Sharma Solutions for Class 9 Maths Chapter 4 Exercise 4.3?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-4-exercise-4-3\/#from-where-can-i-download-the-pdf-of-rd-sharma-solutions-class-9-maths-chapter-4-exercise-43\" title=\"From where can I download the PDF of RD Sharma Solutions Class 9 Maths Chapter 4 Exercise 4.3?\">From where can I download the PDF of RD Sharma Solutions Class 9 Maths Chapter 4 Exercise 4.3?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-class-9-solutions-chapter-4-exercise-43-pdf\"><\/span><strong>Download RD Sharma Class 9 Solutions Chapter 4 Exercise 4.3 PDF<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-SHARMA-Solutions-Class-9-Maths-Chapter-4-Ex-4.3-1.pdf\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Solutions Chapter 4 Exercise 4.3<\/a><\/p>\n<div id=\"example1\" style=\"text-align: justify;\">\u00a0<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 500px;}<br \/>\n.pdfobject { border: 1px solid #666; }<br \/>\n<\/style><\/p>\n<p style=\"text-align: justify;\"><script src=\"https:\/\/www.kopykitab.com\/_utility\/js\/pdfobject.min.js\"><\/script><br \/><script>PDFObject.embed(\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-SHARMA-Solutions-Class-9-Maths-Chapter-4-Ex-4.3-1.pdf\", \"#example1\");<\/script><\/p>\n<h2><span class=\"ez-toc-section\" id=\"access-rd-sharma-class-9-solutions-chapter-4-exercise-43\"><\/span><strong>Access RD Sharma Class 9 Solutions Chapter 4 Exercise 4.3<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><strong>Question 1: Find the cube of each of the following binomial expressions:<\/strong><\/p>\n<p><strong>(i) (1\/x + y\/3)<\/strong><\/p>\n<p><strong>(ii) (3\/x \u2013 2\/x<sup>2<\/sup>)<\/strong><\/p>\n<p><strong>(iii) (2x + 3\/x)<\/strong><\/p>\n<p><strong>(iv) (4 \u2013 1\/3x)<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>[Using identities: (a + b)<sup>3<\/sup>\u00a0= a<sup>3<\/sup>\u00a0+ b<sup>3<\/sup>\u00a0+ 3ab(a + b) and (a \u2013 b)<sup>3<\/sup>\u00a0= a<sup>3<\/sup>\u00a0\u2013 b<sup>3<\/sup>\u00a0\u2013 3ab(a \u2013 b) ]<\/p>\n<p><strong>(i)<\/strong><\/p>\n<p><img title=\"RD sharma class 9 maths chapter 4 ex 4.3 Q1 part 1 solution\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-4-ex-4-3-q1-part-1.png\" alt=\"RD sharma class 9 maths chapter 4 ex 4.3 Q1 part 1 solution\" \/><\/p>\n<p><strong>(ii)<\/strong><\/p>\n<p><img title=\"RD sharma class 9 maths chapter 4 ex 4.3 Q1 part 2 solution\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-4-ex-4-3-q1-part-2.png\" alt=\"RD sharma class 9 maths chapter 4 ex 4.3 Q1 part 2 solution\" \/><\/p>\n<p><strong>(iii)<\/strong><\/p>\n<p><img title=\"RD sharma class 9 maths chapter 4 ex 4.3 Q1 part 3 solution\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-4-ex-4-3-q1-part-3.png\" alt=\"RD sharma class 9 maths chapter 4 ex 4.3 Q1 part 3 solution\" \/><\/p>\n<p><strong>(iv)<\/strong><\/p>\n<p><img title=\"RD sharma class 9 maths chapter 4 ex 4.3 Q1 part 4 solution\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-4-ex-4-3-q1-part-4.png\" alt=\"RD sharma class 9 maths chapter 4 ex 4.3 Q1 part 4 solution\" \/><\/p>\n<p><strong>Question 2: Simplify each of the following:<\/strong><\/p>\n<p><strong>(i) (x + 3)<sup>3<\/sup>\u00a0+ (x \u2013 3)<sup>\u00a03<\/sup><\/strong><\/p>\n<p><strong>(ii) (x\/2 + y\/3)<sup>\u00a03\u00a0<\/sup>\u2013 (x\/2 \u2013 y\/3)<sup>\u00a03<\/sup><\/strong><\/p>\n<p><strong>(iii) (x + 2\/x)<sup>\u00a03<\/sup>\u00a0+ (x \u2013 2\/x)<sup>\u00a03<\/sup><\/strong><\/p>\n<p><strong>(iv) (2x \u2013 5y)<sup>\u00a03\u00a0<\/sup>\u2013 (2x + 5y)<sup>\u00a03<\/sup><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>[Using identities:<\/p>\n<p>a<sup>3<\/sup>\u00a0+ b<sup>3<\/sup>\u00a0= (a + b)(a<sup>2<\/sup>\u00a0+ b<sup>2<\/sup>\u00a0\u2013 ab)<\/p>\n<p>a<sup>3<\/sup>\u00a0\u2013 b<sup>3<\/sup>\u00a0= (a \u2013 b)(a<sup>2<\/sup>\u00a0+ b<sup>2<\/sup>\u00a0+ ab)<\/p>\n<p>(a + b)(a-b) = a<sup>2<\/sup>\u00a0\u2013 b<sup>2<\/sup><\/p>\n<p>(a + b)<sup>2\u00a0<\/sup>= a<sup>2<\/sup>\u00a0+ b<sup>2<\/sup>\u00a0+ 2ab and<\/p>\n<p>(a \u2013 b)<sup>2\u00a0<\/sup>= a<sup>2<\/sup>\u00a0+ b<sup>2<\/sup>\u00a0\u2013 2ab]<\/p>\n<p><strong>(i)<\/strong>\u00a0(x + 3)<sup>3<\/sup>\u00a0+ (x \u2013 3)<sup>\u00a03<\/sup><\/p>\n<p>Here a = (x + 3), b = (x \u2013 3)<\/p>\n<p><img title=\"RD sharma class 9 maths chapter 4 ex 4.3 Q2 solution part 1\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/10\/rd-sharma-class-9-maths-chapter-4-ex-4-3-q2-soluti.png\" alt=\"RD sharma class 9 maths chapter 4 ex 4.3 Q2 solution part 1\" \/><\/p>\n<p><strong>(ii)<\/strong>\u00a0(x\/2 + y\/3)<sup>\u00a03\u00a0<\/sup>\u2013 (x\/2 \u2013 y\/3)<sup>\u00a03<\/sup><\/p>\n<p>Here a = (x\/2 + y\/3) and b = (x\/2 \u2013 y\/3)<\/p>\n<p><img title=\"RD sharma class 9 maths chapter 4 ex 4.3 Q2 solution part 2\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-4-ex-4-3-q2-soluti-1.png\" alt=\"RD sharma class 9 maths chapter 4 ex 4.3 Q2 solution part 2\" \/><\/p>\n<p><strong>(iii)<\/strong>\u00a0(x + 2\/x)<sup>\u00a03<\/sup>\u00a0+ (x \u2013 2\/x)<sup>\u00a03<\/sup><\/p>\n<p>Here a = (x + 2\/x) and b = (x \u2013 2\/x)<\/p>\n<p><sup><img title=\"RD sharma class 9 maths chapter 4 ex 4.3 Q2 solution part 3\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-4-ex-4-3-q2-soluti-2.png\" alt=\"RD sharma class 9 maths chapter 4 ex 4.3 Q2 solution part 3\" \/><\/sup><\/p>\n<p><strong>(iv)<\/strong>\u00a0(2x \u2013 5y)<sup>\u00a03\u00a0<\/sup>\u2013 (2x + 5y)<sup>\u00a03<\/sup><\/p>\n<p>Here a = (2x \u2013 5y) and b = 2x + 5y<\/p>\n<p><img title=\"RD sharma class 9 maths chapter 4 ex 4.3 Q2 solution part 4\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-4-ex-4-3-q2-soluti-3.png\" alt=\"RD sharma class 9 maths chapter 4 ex 4.3 Q2 solution part 4\" \/><\/p>\n<p><strong>Question 3: If a + b = 10 and ab = 21, find the value of a<sup>3<\/sup>\u00a0+ b<sup>3<\/sup>.<\/strong><\/p>\n<p><strong>Solution<\/strong>:<\/p>\n<p>a + b = 10, ab = 21 (given)<\/p>\n<p>Choose a + b = 10<\/p>\n<p>Cubing both sides,<\/p>\n<p>(a + b)<sup>3<\/sup>\u00a0= (10)<sup>3<\/sup><\/p>\n<p>a<sup>3<\/sup>\u00a0+ b<sup>3<\/sup>\u00a0+ 3ab(a + b) = 1000<\/p>\n<p>a<sup>3<\/sup>\u00a0+ b<sup>3<\/sup>\u00a0+ 3 x 21 x 10 = 1000 (using given values)<\/p>\n<p>a<sup>3<\/sup>\u00a0+ b<sup>3<\/sup>\u00a0+ 630 = 1000<\/p>\n<p>a<sup>3<\/sup>\u00a0+ b<sup>3<\/sup>\u00a0= 1000 \u2013 630 = 370<\/p>\n<p>or a<sup>3<\/sup>\u00a0+ b<sup>3<\/sup>\u00a0= 370<\/p>\n<p><strong>Question 4: If a \u2013 b = 4 and ab = 21, find the value of a<sup>3<\/sup>\u00a0\u2013 b<sup>3<\/sup>.<\/strong><\/p>\n<p><strong>Solution<\/strong>:<\/p>\n<p>a \u2013 b = 4, ab= 21 (given)<\/p>\n<p>Choose a \u2013 b = 4<\/p>\n<p>Cubing both sides,<\/p>\n<p>(a \u2013 b)<sup>3<\/sup>\u00a0= (4)<sup>3<\/sup><\/p>\n<p>a<sup>3<\/sup>\u00a0\u2013 b<sup>3<\/sup>\u00a0\u2013 3ab (a \u2013 b) = 64<\/p>\n<p>a<sup>3<\/sup>\u00a0\u2013 b<sup>3<\/sup>\u00a0\u2013 3 \u00d7 21 x 4 = 64 (using given values)<\/p>\n<p>a<sup>3<\/sup>\u00a0\u2013 b<sup>3<\/sup>\u00a0\u2013 252 = 64<\/p>\n<p>a<sup>3<\/sup>\u00a0\u2013 b<sup>3<\/sup>\u00a0= 64 + 252<\/p>\n<p>= 316<\/p>\n<p>Or a<sup>3<\/sup>\u00a0\u2013 b<sup>3<\/sup>\u00a0= 316<\/p>\n<p><strong>Question 5: If x + 1\/x = 5, find the value of x<sup>3<\/sup>\u00a0+ 1\/x<sup>3\u00a0<\/sup>.<\/strong><\/p>\n<p><strong>Solution<\/strong>:<\/p>\n<p>Given: x + 1\/x = 5<\/p>\n<p>Apply Cube on x + 1\/x<\/p>\n<p><img title=\"RD sharma class 9 maths chapter 4 ex 4.3 Q5 solution\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-4-ex-4-3-q5-soluti.png\" alt=\"RD sharma class 9 maths chapter 4 ex 4.3 Q5 solution\" \/><\/p>\n<p><strong>Question 6: If x \u2013 1\/x = 7, find the value of x<sup>3<\/sup>\u00a0\u2013 1\/x<sup>3\u00a0<\/sup>.<\/strong><\/p>\n<p><strong>Solution<\/strong>:<\/p>\n<p>Given: x \u2013 1\/x = 7<\/p>\n<p>Apply Cube on x \u2013 1\/x<\/p>\n<p><img title=\"RD sharma class 9 maths chapter 4 ex 4.3 Q6 solution\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-4-ex-4-3-q6-soluti.png\" alt=\"RD sharma class 9 maths chapter 4 ex 4.3 Q6 solution\" \/><\/p>\n<p><strong>Question 7: If x \u2013 1\/x = 5, find the value of x<sup>3<\/sup>\u00a0\u2013 1\/x<sup>3\u00a0<\/sup>.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: x \u2013 1\/x = 5<\/p>\n<p>Apply Cube on x \u2013 1\/x<\/p>\n<p><strong><img title=\"RD sharma class 9 maths chapter 4 ex 4.3 Ques 7 solution\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-4-ex-4-3-ques-7-so.png\" alt=\"RD sharma class 9 maths chapter 4 ex 4.3 Ques 7 solution\" \/><\/strong><\/p>\n<p><strong>Question 8: If (x<sup>2<\/sup>\u00a0+ 1\/x<sup>2<\/sup>) = 51, find the value of x<sup>3<\/sup>\u00a0\u2013 1\/x<sup>3<\/sup>.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We know that: (x \u2013 y)<sup>2<\/sup>\u00a0= x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0\u2013 2xy<\/p>\n<p>Replace y with 1\/x, we get<\/p>\n<p>(x \u2013 1\/x)<sup>2<\/sup>\u00a0= x<sup>2<\/sup>\u00a0+ 1\/x<sup>2<\/sup>\u00a0\u2013 2<\/p>\n<p>Since (x<sup>2<\/sup>\u00a0+ 1\/x<sup>2<\/sup>) = 51 (given)<\/p>\n<p>(x \u2013 1\/x)<sup>2<\/sup>\u00a0= 51 \u2013 2 = 49<\/p>\n<p>or (x \u2013 1\/x) = \u00b17<\/p>\n<p>Now, Find x<sup>3<\/sup>\u00a0\u2013 1\/x<sup>3<\/sup><\/p>\n<p>We know that, x<sup>3<\/sup>\u00a0\u2013 y<sup>3<\/sup>\u00a0= (x \u2013 y)(x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0+ xy)<\/p>\n<p>Replace y with 1\/x, we get<\/p>\n<p>x<sup>3<\/sup>\u00a0\u2013 1\/x<sup>3<\/sup>\u00a0= (x \u2013 1\/x)(x<sup>2<\/sup>\u00a0+ 1\/x<sup>2<\/sup>\u00a0+ 1)<\/p>\n<p>Use (x \u2013 1\/x) = 7 and (x<sup>2<\/sup>\u00a0+ 1\/x<sup>2<\/sup>) = 51<\/p>\n<p>x<sup>3<\/sup>\u00a0\u2013 1\/x<sup>3<\/sup>\u00a0= 7 x 52 = 364<\/p>\n<p>x<sup>3<\/sup>\u00a0\u2013 1\/x<sup>3<\/sup>\u00a0= 364<\/p>\n<p><strong>Question 9: If (x<sup>2<\/sup>\u00a0+ 1\/x<sup>2<\/sup>) = 98, find the value of x<sup>3<\/sup>\u00a0+ 1\/x<sup>3<\/sup>.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We know that: (x + y)<sup>2<\/sup>\u00a0= x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0+ 2xy<\/p>\n<p>Replace y with 1\/x, we get<\/p>\n<p>(x + 1\/x)<sup>2<\/sup>\u00a0= x<sup>2<\/sup>\u00a0+ 1\/x<sup>2<\/sup>\u00a0+ 2<\/p>\n<p>Since (x<sup>2<\/sup>\u00a0+ 1\/x<sup>2<\/sup>) = 98 (given)<\/p>\n<p>(x + 1\/x)<sup>2<\/sup>\u00a0= 98 + 2 = 100<\/p>\n<p>or (x + 1\/x) = \u00b110<\/p>\n<p>Now, Find x<sup>3<\/sup>\u00a0+ 1\/x<sup>3<\/sup><\/p>\n<p>We know that, x<sup>3<\/sup>\u00a0+ y<sup>3<\/sup>\u00a0= (x + y)(x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0\u2013 xy)<\/p>\n<p>Replace y with 1\/x, we get<\/p>\n<p>x<sup>3<\/sup>\u00a0+ 1\/x<sup>3<\/sup>\u00a0= (x + 1\/x)(x<sup>2<\/sup>\u00a0+ 1\/x<sup>2<\/sup>\u00a0\u2013 1)<\/p>\n<p>Use (x + 1\/x) = 10 and (x<sup>2<\/sup>\u00a0+ 1\/x<sup>2<\/sup>) = 98<\/p>\n<p>x<sup>3<\/sup>\u00a0+ 1\/x<sup>3<\/sup>\u00a0= 10 x 97 = 970<\/p>\n<p>x<sup>3<\/sup>\u00a0+ 1\/x<sup>3<\/sup>\u00a0= 970<\/p>\n<p><strong>Question 10: If 2x + 3y = 13 and xy = 6, find the value of 8x<sup>3<\/sup>\u00a0+ 27y<sup>3<\/sup>.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: 2x + 3y = 13, xy = 6<\/p>\n<p>Cubing 2x + 3y = 13 both sides, we get<\/p>\n<p>(2x + 3y)<sup>3<\/sup>\u00a0= (13)<sup>3<\/sup><\/p>\n<p>(2x)<sup>3<\/sup>\u00a0+ (3y)\u00a0<sup>3<\/sup>\u00a0+ 3( 2x )(3y) (2x + 3y) = 2197<\/p>\n<p>8x<sup>3<\/sup>\u00a0+ 27y<sup>3<\/sup>\u00a0+ 18xy(2x + 3y) = 2197<\/p>\n<p>8x<sup>3<\/sup>\u00a0+ 27y<sup>3<\/sup>\u00a0+ 18 x 6 x 13 = 2197<\/p>\n<p>8x<sup>3<\/sup>\u00a0+ 27y<sup>3<\/sup>\u00a0+ 1404 = 2197<\/p>\n<p>8x<sup>3<\/sup>\u00a0+ 27y<sup>3<\/sup>\u00a0= 2197 \u2013 1404 = 793<\/p>\n<p>8x<sup>3<\/sup>\u00a0+ 27y<sup>3<\/sup>\u00a0= 793<\/p>\n<p><strong>Question 11: If 3x \u2013 2y= 11 and xy = 12, find the value of 27x<sup>3<\/sup>\u00a0\u2013 8y<sup>3<\/sup>.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: 3x \u2013 2y = 11 and xy = 12<\/p>\n<p>Cubing 3x \u2013 2y = 11 both sides, we get<\/p>\n<p>(3x \u2013 2y)<sup>3<\/sup>\u00a0= (11)<sup>3<\/sup><\/p>\n<p>(3x)<sup>3<\/sup>\u00a0\u2013 (2y)<sup>3<\/sup>\u00a0\u2013 3 ( 3x)( 2y) (3x \u2013 2y) =1331<\/p>\n<p>27x<sup>3<\/sup>\u00a0\u2013 8y<sup>3<\/sup>\u00a0\u2013 18xy(3x -2y) =1331<\/p>\n<p>27x<sup>3<\/sup>\u00a0\u2013 8y<sup>3<\/sup>\u00a0\u2013 18 x 12 x 11 = 1331<\/p>\n<p>27x<sup>3<\/sup>\u00a0\u2013 8y<sup>3<\/sup>\u00a0\u2013 2376 = 1331<\/p>\n<p>27x<sup>3<\/sup>\u00a0\u2013 8y<sup>3<\/sup>\u00a0= 1331 + 2376 = 3707<\/p>\n<p>27x<sup>3<\/sup>\u00a0\u2013 8y<sup>3<\/sup>\u00a0= 3707<\/p>\n<p>This is the complete blog of RD Sharma Solutions Class 9 Maths Chapter 4 Exercise 4.3. To know more about the <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\">CBSE<\/a> Class 9 Maths, ask in the comments.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-class-9-solutions-chapter-4-exercise-43\"><\/span><strong>FAQs on RD Sharma Class 9 Solutions Chapter 4 Exercise 4.3<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1631033197266\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-much-does-it-cost-to-download-the-pdf-of-rd-sharma-solutions-for-class-9-maths-chapter-4-exercise-43\"><\/span>How much does it cost to download the PDF of RD Sharma Solutions for Class 9 Maths Chapter 4 Exercise 4.3?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can download it for free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631033208059\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-many-questions-are-there-in-rd-sharma-solutions-for-class-9-maths-chapter-4-exercise-43\"><\/span>How many questions are there in RD Sharma Solutions for Class 9 Maths Chapter 4 Exercise 4.3?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>There are 19 questions in\u00a0RD Sharma Solutions Class 9 Maths Chapter 4 Exercise 4.3.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631033248730\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"from-where-can-i-download-the-pdf-of-rd-sharma-solutions-class-9-maths-chapter-4-exercise-43\"><\/span>From where can I download the PDF of RD Sharma Solutions Class 9 Maths Chapter 4 Exercise 4.3?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can find the download link from the above blog.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Class 9 Solutions Chapter 4 Exercise 4.3: Clear your exams with the RD Sharma Solutions Class 9 Maths. You can easily download the PDF of RD Sharma Class 9 Solutions Chapter 4 Exercise 4.3. All the solutions are as per the current CBSE syllabus and are easy to understand. To know more, read &#8230; <a title=\"RD Sharma Class 9 Solutions Chapter 4 Exercise 4.3 (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-4-exercise-4-3\/\" aria-label=\"More on RD Sharma Class 9 Solutions Chapter 4 Exercise 4.3 (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":243,"featured_media":125230,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411],"tags":[3086,4388],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125014"}],"collection":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/users\/243"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=125014"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125014\/revisions"}],"predecessor-version":[{"id":499623,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125014\/revisions\/499623"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/125230"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=125014"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=125014"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=125014"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}