Câu hỏi của Đoàn Thế Đan

Question

Cho a, b, c thỏa mãn abc=1. Tính A=(a+b+c)³ -(b+c-a)³- (c+a-b)³-(a+b-c)³.

Đoàn Đức Hà · Accepted Answer

$A=\left(a+b+c\right)^3-\left(b+c-a\right)^3-\left(c+a-b\right)^3-\left(a+b-c\right)^3$

Đặt $x=b+c-a,y=c+a-b,z=a+b-c$.

Khi đó $x+y+z=a+b+c$.

$A=\left(x+y+z\right)^3-x^3-y^3-z^3$

$=\left(y+z\right)\left[\left(x+y+z\right)^2+x\left(x+y+z\right)+x^2\right]-\left(y+z\right)\left(y^2-yz+z^2\right)$

$=\left(y+z\right)\left(3x^2+3xy+3zx+3yz\right)$

$=3\left(x+y\right)\left(y+z\right)\left(z+x\right)$

$=24abc=24$

Câu trả lời:

$A=\left(a+b+c\right)^3-\left(b+c-a\right)^3-\left(c+a-b\right)^3-\left(a+b-c\right)^3$

Đặt $x=b+c-a,y=c+a-b,z=a+b-c$.

Khi đó $x+y+z=a+b+c$.

$A=\left(x+y+z\right)^3-x^3-y^3-z^3$

$=\left(y+z\right)\left[\left(x+y+z\right)^2+x\left(x+y+z\right)+x^2\right]-\left(y+z\right)\left(y^2-yz+z^2\right)$

$=\left(y+z\right)\left(3x^2+3xy+3zx+3yz\right)$

$=3\left(x+y\right)\left(y+z\right)\left(z+x\right)$

$=24abc=24$

Yen Nhi · Answer

Đặt $\hept{\begin{cases}a+b-c=x\b+c-a=y\c+a-b=n\end{cases}}$

$\Rightarrow x+y+n=a+b+c$

$\Rightarrow x+y+2b$

$\Rightarrow y+n=2c$

$\Rightarrow n+x=2a$

Ta có:

$A=\left(x+y+n\right)^3-y^3-n^3-x^3$

$=x^3+y^3+n^3+3\left(x+y\right)\left(y+n\right)\left(n+x\right)-y^3-n^3-x^3$

$=3\left(x+y\right)\left(y+n\right)\left(n+z\right)$

$=3.2b.2c.2a=24abc$

$\Rightarrow A=24$ (Vì đề ra $abc=1$)