X+Y+Z=0 THEN FIND VALUE OF X (CUBE)+Y (CUBE)+Z (CUBE)=3XYZ
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If x+y+z=0 then x^3+y^3+z^3=3xyz
dibakartaj:
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Answered by
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We know that,
{a³+b³+c³-3abc = (a+b+c)(a²+b²+c²-ab-bc-ac)}
Apply this formula here,
X³+Y³+Z³ – 3XYZ = (X+Y+Z) (X²+Y²+Z²-XY-YZ-XZ)
Given (X+Y+Z=0)
X³+Y³+Z³ – 3XYZ = (0) (X²+Y²+Z²-XY-YZ-XZ)
X³+Y³+Z³ – 3XYZ = 0
X³+Y³+Z³ = 3XYZ
Hence proved.
Hope it helps...
{a³+b³+c³-3abc = (a+b+c)(a²+b²+c²-ab-bc-ac)}
Apply this formula here,
X³+Y³+Z³ – 3XYZ = (X+Y+Z) (X²+Y²+Z²-XY-YZ-XZ)
Given (X+Y+Z=0)
X³+Y³+Z³ – 3XYZ = (0) (X²+Y²+Z²-XY-YZ-XZ)
X³+Y³+Z³ – 3XYZ = 0
X³+Y³+Z³ = 3XYZ
Hence proved.
Hope it helps...
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