Question

Cho a+b=-1;a.b=-6 Tính P=a^3+b^3

Answer 1

\(P=a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)=\left(-1\right)^3-3\left(-6\right)\left(-1\right)=-1-18=-19\)

Answer 2

\(a+b=-1\Rightarrow\left(a+b\right)^2=1\)

\(\Rightarrow a^2+2ab+b^2=1\Rightarrow a^2+b^2=1-2ab=1-2.\left(-6\right)=13\)

\(P=a^3+b^3=\left(a+b\right)\left(a^2+b^2-ab\right)=\left(-1\right).\left(13+6\right)=-19\)

Answer 3

Ta có: \(\left(a+b\right)^2=\left(-1\right)^2\)

        \(\Rightarrow a^2+2ab+b^2=1\)

        \(\Rightarrow a^2+b^2=1-2ab\)

        \(\Rightarrow a^2+b^2=1-2.\left(-6\right)\)

        \(\Rightarrow a^2+b^2=1+12=13\)

P = \(a^3+b^3=\left(a+b\right).\left(a^2-2ab+b^2\right)=\left(-1\right).\left[13-2.\left(-6\right)\right]=\left(-1\right).\left(25\right)=-25\)