Question

\(Cho\) \(x+y=5\) và \(x^2+y^2=27.\) Tính \(x^3+y^3\)

Answer 1

Cách 2 :

\(x+y=5\\ \Rightarrow\left(x+y\right)^2=25\\ \Rightarrow x^2+y^2+2xy=25\\ \Rightarrow2xy=2\\ \Rightarrow xy=1\)

\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)\\ =5^3-3.1.5\\ =125-15=110\)

Answer 2

Ta có:

x²+y²=27

(x+y)²-2xy=27

25-2xy=27

xy=-1

Ta có:

x+y=5

x²+y²=27

=>(x+y)(x²+y²)=135

x³+y³+x²y+xy²=135

x³+y³+xy(x+y)=135

x³+y³-5=135

x³+y³=140

Answer 3

ta có : \(x^2+y^2=\left(x+y\right)^2-2xy=25-2xy=27\Leftrightarrow xy=-1\)

\(\Rightarrow x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)=5\left(27+1\right)=140\)

vậy \(x^3+y^3=140\)