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Yogita Ingle 6 years, 7 months ago
If x+y+z=0, show that x3 + y3 + z3 = 3xyz
Given that :
x + y + z = 0 ......(1)
We know that,
x3 + y3 + z3 - 3xyz =(x + y + z)(x2 + y2 + z2 - xy - yz - zx) ......(2)
Substituting the value of (1) in (2), we get
x3 + y3 + z3 - 3xyz = 0
x3 + y3 + z3 = 3xyz
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Gaurav Seth 6 years, 7 months ago
LHS = (5 + 2√3 ) / ( 7 + 4√3 )
rationalize the denominator
= (5 + 2√3 ) ( 7 - 4√3 ) / [ ( 7 + 4√3 ) ( 7 - 4√3 )
= [ 5 ×7 - 5 × 4√3 + 2√3 × 7 - 2√3 × 4√3 ] / [ (7 )² - (4√3 )² ]
here we used ( x + y ) (x - y ) = x² - y² identity
= [35 -20√3 + 14√3 -24 ] / [ 49 - 48 ]
= (11 - 6√3 )
therefore ,
LHS = RHS
11 - 6√3 = A - B√3
comparing both sides
A = 11,
B = 6
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