factorize:-250(x-y)³+2
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Answered by
2
Given,
250(x-y)³+2
Calculation,
=250(x-y)³+2
=2(125(x-y)³+1)[We take 2 as common]
Using a³+b³=(a+b)(a²-ab+b²),
=2(5(x-y)+1×(25(x-y)²-5(x-y)+1
=2(5x-5y+1)×(25(x-y)²-5x-5y+1
Answer=>2(5x-5y+1)×(25(x-y)²-5x-5y+1
250(x-y)³+2
Calculation,
=250(x-y)³+2
=2(125(x-y)³+1)[We take 2 as common]
Using a³+b³=(a+b)(a²-ab+b²),
=2(5(x-y)+1×(25(x-y)²-5(x-y)+1
=2(5x-5y+1)×(25(x-y)²-5x-5y+1
Answer=>2(5x-5y+1)×(25(x-y)²-5x-5y+1
Anonymous:
thanks but my question was different ......... it is 250(x-y)³-2 not 1 as u did in the solution
Answered by
2
Hey there !!!!!
=250(x-y)³+2
=2*125(x-y)³+2-----Equation 1
In equation 1 taking "2" common
=2(125(x-y)³+1)
But 5*5*5=5³=125
So,
=2(5³(x-y)³+1)-----Equation 2
In equation 2
(x-y)³ is of the form (a - b)³ = a³ - 3a²b + 3ab² - b³
Here a=x and b=y
So (x-y)³=x³-3x²y+3xy²-y³
Equation 2 changes to
=2(5³(x³-3x²y+3xy²-y³)+1)
=2(125x³-125*3x²y-125y³+1)
=2(125x³-375x²y-125y³+1)
=250(x-y)³+2
=2*125(x-y)³+2-----Equation 1
In equation 1 taking "2" common
=2(125(x-y)³+1)
But 5*5*5=5³=125
So,
=2(5³(x-y)³+1)-----Equation 2
In equation 2
(x-y)³ is of the form (a - b)³ = a³ - 3a²b + 3ab² - b³
Here a=x and b=y
So (x-y)³=x³-3x²y+3xy²-y³
Equation 2 changes to
=2(5³(x³-3x²y+3xy²-y³)+1)
=2(125x³-125*3x²y-125y³+1)
=2(125x³-375x²y-125y³+1)
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