IF x = 2-y, then show that x³+6xy+y³-8=0
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okk... 9th ??
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Answers
Answered by
7
Given,
x = 2-y
=> x + y = 2
cubing on both sides
= > x³+y³+3xy(x+y) = 8
x³ + y³ + 3xy ( 2 ) = 8
since x+y =2
=> x³ + y³ + 6xy = 8
x³ + y³ + 6xy - 8 = 0
x = 2-y
=> x + y = 2
cubing on both sides
= > x³+y³+3xy(x+y) = 8
x³ + y³ + 3xy ( 2 ) = 8
since x+y =2
=> x³ + y³ + 6xy = 8
x³ + y³ + 6xy - 8 = 0
Answered by
3
Given x=2-y
x+y-2=0
Now x³+6xy+y³-8
=x³+y³+(-2)³-3xy(-2)
=[x+y+(-2)] [x²+y²+(-2)²-xy-x(-2)-y(-2)]
=(x+y-2) (x²+y²+4-xy+2x+2y)
=0 (x²+y²+4-xy+2x+2y)
=0
x+y-2=0
Now x³+6xy+y³-8
=x³+y³+(-2)³-3xy(-2)
=[x+y+(-2)] [x²+y²+(-2)²-xy-x(-2)-y(-2)]
=(x+y-2) (x²+y²+4-xy+2x+2y)
=0 (x²+y²+4-xy+2x+2y)
=0
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