Verify that x³ + y³ + z³ - 3xyz = 1/2 (x + y + z)[(x - y)² + (y - z)² + (z - x)²]
Solution:
Taking the RHS and evaluating,
R.H.S. = 1/2 ( x + y + z) [(x - y)² + ( y - z)² + (z - x)²]
= 1/2 ( x + y + z) [(x² - 2xy + y²) + ( y² - 2yz + z²) + (z² - 2zx + x²)]
= 1/2 ( x + y + z) [2x² + 2y² + 2z² - 2xy - 2yz - 2zx]
= 1/2 ( x + y + z) (2) [x² + y² + z² - xy - yz - zx]
= x[x² + y² + z² - xy - yz - zx] + y[x² + y² + z² - xy - yz - zx] + z[x² + y² + z² - xy - yz - zx]
= x³ + xy² + xz² - x²y - xyz - x²z + x²y + y³ + yz² - xy² - y²z - xyz + zx² + y²z + z³ - xyz - yz² - xz²
On simplifying,
= x³ + y³ + z³ - 3xyz = LHS
☛ Check: NCERT Solutions Class 9 Maths Chapter 2
Video Solution:
Verify that x³ + y³ + z³ - 3xy = 1/2 (x + y + z)[(x - y)² + (y - z)² + (z - x)²]
NCERT Solutions Class 9 Maths Chapter 2 Exercise 2.5 Question 12
Summary:
Hence, it is verified that x³ + y³ + z³ − 3xyz = 1/2 (x + y + z) [(x − y)² + (y − z)² + (z − x)²]
☛ Related Questions:
- Factorise each of the following:i) 8a³ + b³ + 12a²b + 6ab²ii) 8a³ - b³ - 12a²b + 6ab²iii) 27 - 125a³ - 135a + 225a²iv) 64a³ - 27b³ - 144a²b + 108ab²v) 27p³ - 1/216 - 9/2p² + 1/4p
- Verify:i) (x³ + y³) = (x + y)(x² - xy + y²)ii) (x³ - y³) = (x - y)(x² + xy + y²)
- Factorise each of the following:i) 27y³ + 125z³ii) 64m³ - 343n³ = (4m)³ - (7n)³[Hint: See Question 9.]
- Factorise: 27x³ + y³ + z³ - 9xyz
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