factorice (x-y)^3+(y-z)^3+(z-x)^3
Answers
(x-y)^3+(y-z)^3+(z-x)^3
=(x-y)(x^2+xy +y^2)+(y-z)(y^2+yz+z^2)+(z-x)(z^2+xz+x^2)
Step-by-step explanation:
please mark me as a brain list......
Answer:
3 [ x²(z - y) + y²(x - z) + z²(y - x) ]
Step-by-step explanation:
Given : (x - y)³ + (y - z)³ + (z - x)³
• Identity : (a - b)³ = a³ - b³ - 3a²b + 3ab²
→ [ (x)³ - (y)³ - 3(x)²(y) + 3(x)(y)² ] + [ (y)³ - (z)³ - 3(y)²(z) + 3(y)(z)² ] + [ (z)³ - (x)³ - 3(z)²(x) + 3(z)(x)² ]
→ [ x³ - y³ - 3x²y + 3xy² ] + [ y³ - z³ - 3y²z + 3yz² ] + [ z³ - x³ - 3z²x + 3zx² ]
- Opening the brackets.
→ x³ - y³ - 3x²y + 3xy² + y³ - z³ - 3y²z + 3yz² + z³ - x³ - 3z²x + 3zx²
- Writing the common terms together.
→ x³ - x³ - y³ + y³ - z³ + z³ + 3zx² - 3x²y + 3xy² - 3y²z + 3yz² - 3z²x
→ 3zx² - 3x²y + 3xy² - 3y²z + 3yz² - 3z²x
We can write this as :
→ 3(zx² - x²y + xy² - y²z + yz² - z²x)
- Taking common terms out.
→ 3 [ x²(z - y) + y²(x - z) + z²(y - x) ]
_______________________________
Or
_______________________________
Given : (x - y)³ + (y - z)³ + (z - x)³
• Identity : a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)
Here, a = x - y, b = y - z, c = z - x
- Finding out (a + b + c).
→ [ (x - y) + (y - z) + (z - x) ]
→ [ x - y + y - z + z - x ]
→ 0
Therefore, a + b + c = 0
Hence,
→ a³ + b³ + c³ - 3abc = (0)(a² + b² + c² - ab - bc - ca)
→ a³ + b³ + c³ - 3abc = 0
→ a³ + b³ + c³ = 3abc ...(1)
Now,
Putting the values of a, b and c in (1).
→ (x - y)³ + (y - z)³ + (z - x)³ = 3 [ (x - y)(y - z)(z - x) ]
Further solving the R.H.S., we get
→ 3 [ x(y - z) - y(y - z) (z - x) ]
→ 3 [ xy - xz - y² + yz (z - x) ]
→ 3 [ z(xy - xz - y² + yz) - x(xy - xz - y² + yz) ]
→ 3 [ xyz - xz² - y²z + yz² - x²y + x²z + xy² - xyz ]
- Writing common terms together.
→ 3 [ xyz - xyz + x²z - x²y + xy² - y²z + yz² - xz² ]
→ 3 [ x²z - x²y + xy² - y²z + yz² - xz² ]
- Taking common terms out.
→ 3 [ x²(z - y) + y²(x - z) + z²(y - x) ]