Question

CMR

(a+b).(b+c).(c+a)=(a+b+c).(ab+bc+ca)-abc

Answer 1

+) \(VT=\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

\(=abc+ac^2+b^2c+bc^2+a^2b+a^2c+ab^2+abc\)

\(=a^2b+a^2c+ab^2+2abc+ac^2+b^2c+bc^2\)

+) \(VP=\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)

\(=a^2b+abc+a^2c+ab^2+b^2c+abc+abc+bc^2+ac^2+\left(-abc\right)\)

\(=a^2b+a^2c+ab^2+2abc+ac^2+b^2c+bc^2\)

\(\Rightarrow VT=VP\) (đpcm)

Answer 2

Ta có : ( a+b) .(b+c) .(c+a) = (ab+ac+b² +bc).(c+a)

= abc+a²b+ac² +a²c+b²c+b²a +bc²+abc

=a²b+abc+ab²+a²c+ c²a+abc +abc +bc² +b²c -abc

=ab.( a+c+b ) + ac.(a+c+b ) +bc.(a+b+c) - abc

= (a+b+c) .(ab+ac+bc) - abc (đpcm)

Answer 3

Ta có:

\(\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

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

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

\(=\left(a+b+c\right)\left(ab+bc+ac\right)-c\left(ab+bc+ac\right)+\left(a+b\right)c^2\)\(=\left(a+b+c\right)\left(ab+bc+ac\right)-abc-bc^2-ac^2+\left(a+b\right)c^2\)\(=\left(a+b+c\right)\left(ab+bc+ac\right)-abc-c^2\left(a+b\right)+\left(a+b\right)c^2\)\(=\left(a+b+c\right)\left(ab+bc+ac\right)-abc\)