Prove that (x+y)3– (x-y)3-6y(x2-y2)=8y3
Answers
Answered by
0
Answer:
=(x+y)
3
−(x−y)
3
−6y(x
2
−y
2
)
=(x+y)
3
−(x−y)
3
−6y(x−y)(x+y)
=(x+y)
3
−(x−y)((x−y)
2
+6y(x+y))
=(x+y)
3
−(x−y)(x
2
+y
2
−2xy+6xy+6y
2
)
=(x+y)
3
−(x−y)(x
2
+7y
2
+4xy)
=x
3
+y
3
+3x
2
y+3xy
2
−x
3
−7xy
2
−4x
2
y+x
2
y+7y
3
+4xy
2
=8y
3
Answered by
1
Step-by-step explanation:
(x+y)³-(x-y)³-6y(x²-y²)=8y³
L.H.S=(x+y-x+y) [(x+y)²+(x+y)(x-y)+(x-y)²]-6y(x²-y²)
= (2y) [( x²+y²+2xy) + x²-y² + (x²+y²-2xy)] -6y(x²-y²)
= (2y) [ x²+y²+2xy+x²-y²+x²+y²-2xy] -6y(x²-y²)
= (2y) [ 3x²+y²] -6x²y+6y³
= 6x²y + 2y³-6x²y+6y³
= 8y³= R.H.S
Hence proved
{ Identities used
a³-b³ = (a-b)(a²+ab+b²)
(a+b)² = a²+b²+2ab
(a-b)²= a²+b²-2ab
(a+b)(a-b)=a²-b²}
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