\(VP=\left(a+b\right)^3-3ab\left(a+b\right)=a^3+3a^2b+3ab^2+b^3-3a^2b-3ab^2=a^3+b^3=VT\)
=> ĐPCM
b; tương tự
\(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\)
= -5^3 - 3.6.-5 = -125 + 90 = - 35
Chứng minh rằng:
a)\(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\)
b)\(\left(a-b\right)^3+3ab\left(a-b\right)=a^3-b^3\)
c)\(\left(a+b\right)^2-\left(a-b\right)^2=4ab\)
Chứng minh
\(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\)
\(\left(a+b+c\right)^3=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
\(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left[\left(a+b\right)+c\right]^3-a^3-b^3-c^3\)
\(=\left(a+b\right)^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3-a^3-b^3-c^3\)
\(=a^3+3a^2b+3ab^2+b^3+3c\left(a^2+2ab+b^2\right)+3ac^2+3bc^2-a^3-b^3\)
\(=3a^2b+3ab^2+3a^2c+6abc+3b^2c+3ac^2+3bc^2\)
\(=3\left(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+2abc\right)\)
\(=3\left[\left(a^2b+ab^2\right)+\left(a^2c+abc\right)+\left(ac^2+bc^2\right)+\left(b^2c+abc\right)\right]\)
\(=3\left[ab\left(a+b\right)+ac\left(a+b\right)+c^2\left(a+b\right)+bc\left(a+b\right)\right]\)
\(=3\left(a+b\right)\left(ab+ac+c^2+bc\right)\)
\(=3\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)
\(=3\left(a+b\right)\left(b+c\right)\left(c+b\right)\)
Chứng minh
\(\left(a+b\right)^3=\left(a-b\right)\left(a^2+ab+b^2\right)-3ab\left(a-b\right)\)
CMR
a) \(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\)
b)\(a^3-b^3=\left(a-b\right)^3+3ab\left(a-b\right)\)
Giúp mk vs mk đg cần gấp
chung minh
\(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\)
\(a^3-b^3=\left(a-b\right)^3+3ab\left(a-b\right)\)
Tính \(M=a^3+b^3+3ab\left(a^2+b^2\right)+6a^2b^2\left(a+b\right)\) biết a+b=1
Cho a + b + c = 6 . Tính giá trị biểu thức :
\(\frac{a^3+b^3+c^3-3ab}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)
Tính:
a, \(N=8a^3-27b^3\)biết ab=12 và 2a-3b=5
b, \(K=a^3+b^3+6a^2b^2\left(a+b\right)+3ab\left(a^2+b^2\right)\)biết a+b=1
c, \(P=\left(\dfrac{x}{4}\right)^3+\left(\dfrac{y}{2}\right)^3\)biết xy=4 và x+2y=8
