Question

Cho x+y=101. Tính giá trị của biểu thức :

P=x³-3x²+3x²y+3xy²+y³-3y³-6xy+3x+3y+2015

Answer 1

\(P=x^3-3x^2+3x^2y+3xy^2+y^3-3y^2-6xy+3x+3y+2015\)

\(\Leftrightarrow P=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(3x^2+6xy+3y^2\right)+\left(3x+3y\right)+2015\)

\(\Leftrightarrow P=\left(x+y\right)^3-3\left(x^2+2xy+y^2\right)+3\left(x+y\right)+2015\)

\(\Leftrightarrow P=\left(x+y\right)^3-3\left(x+y\right)^2+3\left(x+y\right)+2015\)

\(\Leftrightarrow P=101^3-3.101^2+3.101+2015\)

Answer 2

\(P=x^3-3x^2+3x^2y+3xy^2+y^3-3y^2-6xy+3x+3y+2015\)

\(\Leftrightarrow P=x^3+3x^2y+3xy^2+y^3-3x^2-6xy-3y^2+3x+3y+2015\)

\(\Leftrightarrow P=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(3x^2+6xy+3y^2\right)+\left(3x+3y\right)+2015\)

\(\Leftrightarrow P=\left(x^3+3x^2y+3xy^2+y^3\right)-3\left(x^2+2xy+y^2\right)+3\left(x+y\right)+2015\)

\(\Leftrightarrow P=\left(x+y\right)^3-3\left(x+y\right)^2+3\left(x+y\right)+2015\)

\(\Leftrightarrow P=101^3-3.101^2+3.101+2015\)

\(\Leftrightarrow P=1030301-30603+303+2015\)

\(\Leftrightarrow P=999698+303+2015\)

\(\Leftrightarrow P=1000001+2015\)

\(\Leftrightarrow P=1002016\)