Learn all Concepts of Polynomials Class 9 (with VIDEOS). Check - Polynomials Class 9
Last updated at April 10, 2019 by Teachoo
Transcript
Ex 2.5, 14 Without actually calculating the cubes, find the value of each of the following: (-12)3 + (7)3 + (5)3 We know that x3 + y3 + z3 3xyz = (x + y + z) (x2 + y2 + z2 xy yz zx) So, x3 + y3 + z3 = 3xyz + (x + y + z) (x2 + y2 + z2 xy yz zx) Putting x = 12, y = 7 , z = 5 (-12)3 + (7)3 + (5)3 = 3 (-12) (7)(5) + ( 12 + 7 + 5) ((-12)2 + (7)2 + (5)2 (-12)(7) (7)(5) (5)(-12) ) = 3 (-12) (7)(5) + (0) ((-12)2 + (7)2 + (5)2 (-12)(7) (7)(5) (5)(-12) = 3 (-12) (7)(5) = -1260 Ex 2.5, 14 Without actually calculating the cubes, find the value of each of the following: (ii) (28)3 + (-15)3 + (-13)3 We know that x3 + y3 + z3 3xyz = (x + y + z) (x2 + y2 + z2 xy yz zx) So, x3 + y3 + z3 = 3xyz + (x + y + z) (x2 + y2 + z2 xy yz zx) Putting x = 28, y = 15 , z = 13 (28)3 + ( 15)3 + ( 13)3 = 3 (28) ( 15) ( 13) + (28 15 13) ((28)2 +(-15)2+(-13)2 +(28)(-15)+(-15)(-13)+(-13)(28)) = 3(28)( 15)( 13) + 0 = 16380
Ex 2.5
Ex 2.5,2 Important
Ex 2.5,3 Important
Ex 2.5,4 Important
Ex 2.5,5 Important
Ex 2.5,6
Ex 2.5,7 Important
Ex 2.5,8
Ex 2.5,9 Not in Syllabus - CBSE Exams 2021
Ex 2.5,10 Important Not in Syllabus - CBSE Exams 2021
Ex 2.5, 11 Important
Ex 2.5,12 Not in Syllabus - CBSE Exams 2021
Ex 2.5,13 Not in Syllabus - CBSE Exams 2021
Ex 2.5,14 Not in Syllabus - CBSE Exams 2021 You are here
Ex 2.5, 15
Ex 2.5,16 Important
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