(x+y)³-(x-y) ³-6y(x²-y²)
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Given: (x + y)³ - (x - y)³ - 6y(x² - y²) = ky³
In order to factorise the given algebraic expression we use the following Identity - = (a + b)³ = a³ + b³ + 3a²b + 3ab² & (a - b)³ = a³ - b³ - 3a²b + 3ab² :
= x³ + y³ + 3x²y + 3xy² - (x³ - y³ - 3x²y + 3xy²) - 6y(x² - y²) = ky³
= x³ + y³ + 3x²y + 3xy² - x³ + y³ + 3x²y - 3xy² - 6x²y + 6y³ = ky³
= x³ - x³ + y³ + y³ + 6y³ + 3x²y + 3x²y - 6x²y + 3xy² - 3xy² = ky³
= 2y³ + 6y³ = ky²³
= 8y³ = ky³
On comparing :
= k = 8
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Consider,
can be rewritten as
We know,
So, using this algebraic identity, we get
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