Question

chứng minh các hằng đẳng thức

a) (a+b+c)³-a³-b³-c³=3(a+b) (b+c)(c+d)

Answer 1

VT = (a+b+c)³-a³-b³-c³

= \([\left(a+b\right)+c]^3\)- a³-b³-c³

= (a+b)³+c³ +3ab(a+b)+3c(a+b)(a+b+c)-a³-b³-c³

⁼3(a+b) \([ab+c\left(a+b+c\right)]\)

= 3(a+b) \([ab+ac+bc+c^2]\)

= 3(a+b)(b+c)(c+a)

\(\Rightarrow\)VT=VP= 3(a+b)(b+c)(c+a)

Answer 2

Giải:

Ta có: \(VT=\left(a+b+c\right)^3-a^3-b^3-c^3\)

\(=\left[\left(a+b+c\right)^3-a^3\right]-\left(b^3+c^3\right)\)

\(=\left(b+c\right)\left[\left(a+b+c\right)^2+\left(a+b+c\right)a+a^2\right]-\left(b+c\right)\left(b^2-bc+c^2\right)\)

\(=\left(b+c\right)\left(3a^2+3ab+3bc+3ca\right)\)

\(=3\left(b+c\right)\left[a\left(a+b\right)+c\left(a+b\right)\right]\)

\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)=VP\) (Đpcm)