Factorise (x+y–z)³ + (x–y+z)³ – 2x³
Answers
Given :- Factorise (x+y–z)³ + (x–y+z)³ – 2x³
Solution :-
we know that,
→ (a + b + c)³ = a³ + b³ + c³ + 3a²b + 3ab² + 3b²c + 3bc² + 3c²a + 3ca² + 6abc .
solving first part by taking a = x , b = y , c = (-z)
→ (x + y - z)³ = x³ + y³ - z³ + 3x²y + 3xy² - 3y²z + 3yz² + 3z²x - 3zx² - 6xyz
solving second part by taking a = x , b = (-y) , c = z
→ (x - y + z)³ = x³ - y³ + z³ - 3x²y + 3xy² + 3y²z - 3yz² + 3z²x + 3zx² - 6xyz
putting both now we get,
→ (x+y–z)³ + (x–y+z)³ – 2x³
→ [x³ + y³ - z³ + 3x²y + 3xy² - 3y²z + 3yz² + 3z²x - 3zx² - 6xyz] + [x³ - y³ + z³ - 3x²y + 3xy² + 3y²z - 3yz² + 3z²x + 3zx² - 6xyz] - 2x³
→ x³ + x³ - 2x³ + y³ - y³ - z³ + z³ + 3x²y - 3x²y + 3xy² + 3xy² - 3y²z + 3y²z + 3yz² - 3yz² + 3z²x + 3z²x - 3zx² + 3zx² - 6xyz - 6xyz
cancel the like terms now, (x³ + x³ - 2x³ = 2x³ - 2x³)
→ 6xy² + 6z²x - 12xyz
→ 6x(y² + z² - 2yz)
using a² + b² - 2ab = (a - b)² now,
→ 6x(y - z)² or, 6x(z - y)² (Ans.)
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Let a, b and c be non-zero real numbers satisfying (a³)/(b³ + c³) + (b³)/(c³ + a³) + (c³)/(a³ + b³)
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if a²+ab+b²=25
b²+bc+c²=49
c²+ca+a²=64
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Given : (x+y–z)³ + (x–y+z)³ – 2x³
To Find : Factorise
Solution:
(x+y–z)³ + (x–y+z)³ – 2x³
= (x + ( y - z))³ + (x - (y - z))³ - 2x³
(a + b)³ = a³ + b³ + 3ab(a + b)
(a - b)³ = a³ - b³ - 3ab(a - b)
a = x b = (y - z)
= x³ + (y - z)³ + 3x(y - z) (x + y - z) + x³ - (y - z)³ - 3x(y - z) (x - y + z) - 2x³
= 3x(y - z) (x + y - z) - 3x(y - z) (x - y + z)
=3x(y - z)( x + y - z - (x - y + z))
= 3x(y - z)( 2y -2z)
= 3x(y - z)2( y -z)
= 6x(y - z)²
(x+y–z)³ + (x–y+z)³ – 2x³ = 6x(y - z)²
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