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Ex 2.4, 14 (i) - Chapter 2 Class 9 Polynomials
Last updated at April 16, 2024 by Teachoo
Chapter 2 Class 9 Polynomials
Concept wise
- Definition
- Degree & Coefficient of a polynomial
- Value of a polynomial at a given point
- Verifying Zeroes of a polynomial
- Finding Zeroes of a polynomial
- Remainder Theoram
- Check if factor
- Factorisation by middle term
- Factorisation by factor formula
- Factorizing cubic equation
- Identity I - IV
- Identity V
- Identity VI & VII
-
Identity VIII
Transcript
Ex 2.4, 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
