Question 399581
3x(cube)-81y(cube) 
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3[x^3-27y^3]
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= 3[x^3-(3y)^3]
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= 3[x-(3y)][(x^2+3xy+9y^2]
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This is based on the form:
(a^3-b^3) = (a-b)(a^2+ab+b^2)
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Cheers,
Stan H.
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