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Algebra of Derivatives

By how much i...

Question

By how much is $3 x^{3} - 2 x^{2} y + x y^{2} - y^{3}$ less than $4 x^{3} - 3 x^{2} y - 7 x y^{2} + 2 y^{3}$ ?

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Solution

Step: Subtract $(3 x^{3} - 2 x^{2} y + x y^{2} - y^{3})$ from $(4 x^{3} - 3 x^{2} y - 7 x y^{2} + 2 y^{3})$
In order to find the answer, subtract $3 x^{3} - 2 x^{2} y + x y^{2} - y^{3})$ from $(4 x^{3} - 3 x^{2} y - 7 x y^{2} + 2 y^{3})$
That is, $(4 x^{3} - 3 x^{2} y - 7 x y^{2} + 2 y^{3}) - (3 x^{3} - 2 x^{2} y + x y^{2} - y^{3})$
$= 4 x^{3} - 3 x^{2} y - 7 x y^{2} + 2 y^{3} - 3 x^{3} + 2 x^{2} y - x y^{2} + y^{3}$
$= 4 x^{3} - 3 x^{3} - 3 x^{2} y + 2 x^{2} y - 7 x y^{2} - x y^{2} + 2 y^{3} + y^{3}$
$= x^{3} - x^{2} y - 8 x y^{2} + 3 y^{3}$
Hence, $(3 x^{3} - 2 x^{2} y + x y^{2} - y^{3})$ is $(x^{3} - x^{2} y - 8 x y^{2} + 3 y^{3})$ less than $(4 x^{3} - 3 x^{2} y - 7 x y^{2} + 2 y^{3}) .$

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