Factorise:
8(x+y)³+27(x-y) ³
Answers
Answered by
0
Answer:
Step-by-step explanation:
ANSWER
8(x+y)
3
−27(x−y)
3
=[2(x+y)]
3
−[3(x−y)]
3
=[2(x+y)−3(x−y)][(2(x+y))
2
+2(x+y)×3(x−y)+(3(x−y))
2
]
=(2x+2y−3x+3y)[4(x
2
+2xy+y
2
)+6(x
2
−y
2
)+9(x
2
−2xy+y
2
)]
=(−x+5y)(4x
2
+8xy+4y
2
+6x
2
−6y
2
+9x
2
−18xy+9y
2
)
=(−x+5y)(19x
2
−10xy+7y
2
)
Answered by
3
Answer:
8(x+y)³-27(x-y)³
8(x+y)³-27(x-y)³=(-x+5y)(19x²-10xy+7y²)
Step-by-step explanation:
Given 8(x+y)³-27(x-y)³
= [2(x+y)]³-[3(x-y)]³
=(2x+2y)³-(3x-3y)³
We know the algebraic identity:
a³-b³ = (a-b)(a²+ab+b²)
Here ,
a = 2x+2y, b = 3x-3y
=[2x+2y-(3x-3y)][(2x+2y)²+(2x+2y)(3x-3y)+(3x-3y)²]
=(2x+2y-3x+3y)[4x²+8xy+4y²+6x²-6xy+6xy-6y²+9x²-18xy+9y²]
= (-x+5y)(19x²-10xy+7y²)
Therefore,
8(x+y)³-27(x-y)³
=(-x+5y)(19x²-10xy+7y²)
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