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What is the value of 37^3 using the identity (x − y)^3 = x^3 − 3x2y + 3xy^2 − y^3? Hint: 37^3 = (40 − 3)^3; therefore, x = 40 and y = 3.
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@dan815
@shaniehh you have to substitute the values in the formula
\((x − y)^3 = x^3 − 3x2y + 3xy^2 − y^3\)
\((40− 3)^3 = 40^3 − 3*40*2(3) + 3(40)(3)^2 − 3^3\)
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