{"id":337,"date":"2023-10-17T06:22:47","date_gmt":"2023-10-17T00:52:47","guid":{"rendered":"https:\/\/samacheer-kalvi.com\/?p=337"},"modified":"2023-11-10T10:30:17","modified_gmt":"2023-11-10T05:00:17","slug":"samacheer-kalvi-9th-maths-guide-chapter-3-ex-3-4","status":"publish","type":"post","link":"https:\/\/samacheer-kalvi.com\/samacheer-kalvi-9th-maths-guide-chapter-3-ex-3-4\/","title":{"rendered":"Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.4"},"content":{"rendered":"

Students can download Maths Chapter 3 Algebra Ex 3.4 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide<\/a> Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.<\/p>\n

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.4<\/h2>\n

Question 1.
\nExpand the following:
\n(i) (2x + 3y + 4z)2<\/sup>
\n(ii) (-p + 2q + 3r)2<\/sup>
\n(iii) (2p + 3) (2p – 4) (2p – 5)
\n(iv) (3a + 1) (3a – 2) (3a + 4)
\nSolution:
\nWe know that (a + b + c)2<\/sup> = a2<\/sup> + b2<\/sup> + c2<\/sup> + 2ab + 2bc + 2ac
\n(i) (2x + 3y + 4z)2<\/sup> = (2x)2<\/sup> + (3y)2<\/sup> + (4z)2<\/sup> + 2(2x) (3y) + 2(3y) (4z) + 2(4z) (2x)
\n= 4x2<\/sup> + 9y2<\/sup> + 16z2<\/sup> + 12xy + 24yz + 16xz<\/p>\n

(ii) (-p + 2q + 3r)2<\/sup> = (-p)2<\/sup> + (2q)2<\/sup> + (3r)2<\/sup> + 2(-p) (2q) + 2(2q)(3r) + 2(3r) (- p)
\n= p2<\/sup>+ 4q2<\/sup> + 9r2<\/sup> – 4pq + 12qr – 6pr<\/p>\n

\"Samacheer<\/p>\n

(iii) (2p + 3) (2p – 4) (2p – 5)
\n[Here x = 2p, a = 3, b = -4 and c = -5]
\n= (2p)3<\/sup> + (3 – 4 – 5) (2p)2<\/sup> + [(3)(-4) + (-4)(-5) + (3) (-5)] 2p + (3) (-4) (-5)
\n= 8p3<\/sup> + (-6)(4p2<\/sup>) + (-12 + 20 – 15) 2p + 60
\n= 8p3<\/sup> – 24p2<\/sup> – 14p + 60<\/p>\n

(iv) (3a + 1) (3a – 2) (3a + 4)
\n[Here x = 3a, a = 1, b = -2 and c = 4]
\n= (3a)3<\/sup> + (1 – 2 + 4) (3a)2<\/sup> + [(1)(-2) + (-2) (4) + (4) (1)] (3a) + (1) (-2) (4)
\n= 27a3<\/sup> + 3(9a2<\/sup>) + (-2 – 8 + 4) (3a) – 8
\n= 27a3<\/sup> + 27a2<\/sup> – 18a – 8<\/p>\n

\"Samacheer<\/p>\n

Question 2.
\nUsing algebraic identity, find the coefficients of x2<\/sup>, x and constant term without actual expansion.
\n(i) (x + 5)(x + 6)(x + 7)
\nSolution:
\n[Here x = x, a = 5, b = 6, c = 7]
\n(x + a) (x + b) (x + c) = x3<\/sup> + (a + b + c)x2<\/sup> + (ab + bc + ac)x + abc
\ncoefficient of x2<\/sup> = 5 + 6 + 7
\n= 18
\ncoefficient of x = 30 + 42 + 35
\n= 107
\nconstant term = (5) (6) (7)
\n= 210<\/p>\n

\"Samacheer<\/p>\n

(ii) (2x + 3)(2x – 5) (2x – 6)
\nSolution:
\n[Here x = 2x, a = 3, b = -5, c = -6]
\n(x + a) (x + b) (x + c) = x3<\/sup> + (a + b + c)x2<\/sup> + (ab + bc + ac)x + abc
\ncoefficient of x2<\/sup> = (3 – 5 – 6)4 [(2x)2<\/sup> = 4x2<\/sup>]
\n= (-8) (4)
\n= -32
\ncoefficient of x = [(3)(-5) + (-5)(-6) + (-6)(3)](2)
\n= (-15 + 30-18) (2)
\n= (-3) (2)
\n= -6
\nconstant term = (3) (-5) (-6)
\n= 90<\/p>\n

\"Samacheer<\/p>\n

Question 3.
\nIf (x + a)(x + b)(x + c) = x3<\/sup> + 14x2<\/sup> + 59x + 70, find the value of
\n(i) a + b + c
\n(ii) \\(\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}\\)
\n(iii) a2<\/sup> + b2<\/sup> + c2<\/sup>
\n(iv) \\(\\frac{a}{bc} + \\frac{b}{ac} + \\frac{c}{ab}\\)
\nSolution:
\n(x + a) (x + b) (x + c) = x3<\/sup> + 14x2<\/sup> + 59x + 70
\nx3<\/sup> + (a + b + c)x2<\/sup> + (ab + bc + ac)x + abc = x3<\/sup> + 14x2<\/sup> + 59x + 70
\na + b + c = 14, ab + bc + ac = 59, abc = 70
\n(i) a + b + c = 14<\/p>\n

(ii) \\(\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}\\) = \\(\\frac{bc+ac+ab}{abc}\\)
\n= \\(\\frac{59}{70}\\)<\/p>\n

(iii) a2<\/sup> + b2<\/sup> + c2<\/sup> = (a + b + c)2<\/sup> – 2 (ab + bc + ac)
\n= (14)2<\/sup> – 2(59)
\n= 196 – 118
\n= 78<\/p>\n

\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 4.
\nExpand:
\n(i) (3a – 4b)3<\/sup>
\nSolution:
\n(a – b)3<\/sup> = a3<\/sup> – b3<\/sup> – 3ab (a – b)
\n(3a – 4b)3<\/sup> = (3a)3<\/sup> – (4b)3<\/sup> – 3(3a)(4b)(3a – 4b)
\n= 27a3<\/sup> – 64b3<\/sup> – 36ab (3a – 4b)
\n= 27a3<\/sup> – 64b3<\/sup> – 108a2<\/sup>b + 144ab2<\/sup><\/p>\n

(ii) [x + \\(\\frac{1}{y}]^{3}\\)
\nSolution:
\n(a + b)3<\/sup> = a3<\/sup> + b3<\/sup> + 3ab (a + b)
\n\"Samacheer<\/p>\n

Question 5.
\nEvaluate the following by using identities:
\n(i) 983<\/sup>
\nSolution:
\n983<\/sup> = (100 – 2)3<\/sup> [(a – b)3 <\/sup>= a3<\/sup> – b3<\/sup> – 3ab (a – b)]
\n= 1003<\/sup> – (2)3<\/sup> – 3(100) (2) (100 – 2)
\n= 1000000 – 8 – 600(98)
\n= 1000000 – 8 – 58800
\n= 1000000 – 58808
\n= 941192<\/p>\n

(ii) 10013<\/sup>
\nSolution:
\n(1001)3 <\/sup>= (1000 + 1)3<\/sup>
\n[(a + b)3<\/sup> = a3<\/sup> + b3<\/sup> + 3ab (a + b)]
\n= (1000)3<\/sup> + 13<\/sup> + 3(1000) (1) (1000 + 1)
\n= 1000000000 + 1 + 3000 (1001)
\n= 1000000001 + 3003000
\n= 1003003001<\/p>\n

\"Samacheer<\/p>\n

Question 6.
\nIf (x + y + z) = 9 and (xy + yz + zx) = 26, then find the value of x2<\/sup> + y2<\/sup> + z2<\/sup>.
\nSolution:
\nx + y + z = 9; xy + yz + zx = 26
\nx2<\/sup> + y2<\/sup> + z2<\/sup> = (x + y + z)2<\/sup> – 2xy – 2yz – 2xz
\n= (x + y + z)2<\/sup> – 2 (xy + yz + zx)
\n= 92<\/sup> – 2(26)
\n= 81 – 52
\n= 29<\/p>\n

Question 7.
\nFind 27a3<\/sup> + 64b3<\/sup>, If 3a + 4b = 10 and ab = 2
\nSolution:
\n3a + Ab = 10, ab = 2
\n27a3<\/sup> + 64b3<\/sup> = (3a)3<\/sup> + (4b)3<\/sup>
\n[a3<\/sup> + b3<\/sup> = (a + b)3<\/sup> – 3 ab (a + b)]
\n= (3a + 4b)3<\/sup> – 3 \u00d7 3a \u00d7 4b (3a + 4b)
\n= 103 – 36ab (10)
\n= 1000 – 36(2)(10)
\n= 1000 – 720
\n= 280<\/p>\n

\"Samacheer<\/p>\n

Question 8.
\nFind x3<\/sup> – y3<\/sup>, if x – y = 5 and xy = 14.
\nSolution:
\nx – y = 5, xy = 14
\nx3<\/sup> – y3\u00a0 <\/sup>= (x – y)3<\/sup> + 3xy (x – y)
\n= 53<\/sup> + 3(14) (5)
\n= 125 + 210
\n= 335<\/p>\n

Question 9.
\nIf a + \\(\\frac{1}{a}\\) = 6, then find the value of a3<\/sup> +\\(\\frac{1}{a^3}\\)
\nSolution:
\na + \\(\\frac{1}{a}\\) = 6 [a3<\/sup> + b3<\/sup> = (a + b)3<\/sup> – 3ab (a + b)]
\n\"Samacheer
\n= 63<\/sup> – 3(6)
\n= 216 – 18
\n= 198<\/p>\n

\"Samacheer<\/p>\n

Question 10.
\nIf x2<\/sup> + \\(\\frac{1}{x^2}\\) = 23, then find the value of x + \\(\\frac{1}{x}\\) and x3<\/sup> + \\(\\frac{1}{x^3}\\)
\nSolution:
\n\"Samacheer
\nWhen x = 5 [a3<\/sup> + b3<\/sup> = (a + b)3<\/sup> – 3ab (a + b)]
\n= (5)3<\/sup> – 3(5)
\n= 125 – 15
\n= 110
\nwhen x = -5
\nx3<\/sup> + \\(\\frac{1}{x^3}\\) = (-5)3<\/sup> – 3(-5)
\n= -125 + 15
\n= -110
\n\u2234 x3<\/sup> + \\(\\frac{1}{x^3}\\) = \u00b1110<\/p>\n

Question 11.
\nIf (y – \\(\\frac{1}{y})^{3}\\) = 27 then find the value of y3<\/sup> – \\(\\frac{1}{y^3}\\)
\nSolution:
\n\"Samacheer
\n= 33<\/sup> + 3(3)
\n= 27 + 9
\n= 36<\/p>\n

\"Samacheer<\/p>\n

Question 12.
\nSimplify:
\n(i) (2a + 3b + 4c) (4a2<\/sup> + 9b2<\/sup> + 16c2<\/sup> – 6ab – 12bc – 8ca)
\n(ii) (x – 2y + 3z) (x2<\/sup> + 4y2<\/sup> + 9z2<\/sup> + 2xy + 6yz – 3xz)
\nSolution:
\nx3<\/sup> + y3<\/sup> + z3<\/sup> – 3xyz \u2261 (x + y + z) (x2<\/sup> + y2<\/sup> + z2<\/sup> – xy – yz – zx)
\n(i) (2a + 3b + 4c) (4a2<\/sup> + 9b2<\/sup> + 16c2<\/sup>\u00a0– 6ab – 12bc – 8ea)
\n= (2a)3<\/sup> + (3b)3<\/sup> + (4c)3<\/sup> – 3 (2a) (3b) (4c)
\n= 8a3<\/sup> + 27b3<\/sup> + 64c3<\/sup> – 72abc<\/p>\n

(ii) (x – 2y + 3z) (x2<\/sup> + 4y2<\/sup> + 9z2<\/sup> + 2xy + 6yz – 3xz)
\n= x3<\/sup> + (-2y)3<\/sup> + (3z)3<\/sup> – 3(x) (-2y) (3z)
\n= x3<\/sup> – 8y3<\/sup> + 27z3<\/sup> + 18xyz<\/p>\n

Question 13.
\nBy using identity evaluate the following:
\n(i) 73<\/sup> – 103<\/sup> + 33<\/sup>
\nSolution:
\nx3<\/sup> + y3<\/sup> + z3<\/sup> – 3xyz = (x + y + z) (x2<\/sup> + y2<\/sup> + z2<\/sup> – xy – yz – zx)
\nWe know that a + b + c = 0 then a3<\/sup> + b3<\/sup> + c3<\/sup> = 3ab
\na + b + c = 7 + (-10) + 3
\n= 10 – 10
\n= 0
\n\u2234 73<\/sup> – 103<\/sup> + 33<\/sup> = 3(7) (-10) (3)
\n= -630<\/p>\n

(ii) 1 + \\(\\frac{1}{8}\\) – \\(\\frac{27}{8}\\)
\nSolution:
\nWe know that a3<\/sup> + b3<\/sup> + c3<\/sup> = 0 then a + b + c = 3abc
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 14.
\nIf 2x -3y – 4z = 0, then find 8x3<\/sup> – 27y3<\/sup> – 64z3<\/sup>.
\nSolution:
\nWe know x3<\/sup> +y3<\/sup> + z3<\/sup> – 3xyz = (x + y + z) (x2<\/sup> + y2<\/sup> + z2<\/sup> – xy – yz – zx)
\nx3<\/sup> + y3<\/sup> + z3<\/sup> = (x + y + z) (x2<\/sup> +y2<\/sup> + z2<\/sup> – xy – yz – zx) + 3xyz
\n8x3<\/sup> – 27y3<\/sup> – 64z3<\/sup> = (2x)3<\/sup> + (-3y)3<\/sup> + (-4z)3<\/sup>
\n= (2x – 3y- 4z) [(2x)2<\/sup> + (-3y)2<\/sup> + (-4z)2<\/sup> – (2x)(-3y) – (-3y) (-4z) -(-4z)(2x)] + 3(2x)(-3y)(-4z)
\n= 0 (4x2<\/sup> + 9y2<\/sup> + 16z2<\/sup> + 6xy – 12yz + 8xz) + 72xyz
\n= 72xyz
\n8x3<\/sup> – 27y3<\/sup> – 64z3<\/sup> = 72xyz<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can download Maths Chapter 3 Algebra Ex 3.4 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams. Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.4 Question …<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[3],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/posts\/337"}],"collection":[{"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/comments?post=337"}],"version-history":[{"count":1,"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/posts\/337\/revisions"}],"predecessor-version":[{"id":49498,"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/posts\/337\/revisions\/49498"}],"wp:attachment":[{"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/media?parent=337"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/categories?post=337"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheer-kalvi.com\/wp-json\/wp\/v2\/tags?post=337"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}