Question

prove that a + b + C whole cube minus a cube minus b cube minus C cube equals to 3 into a + b into B + c into C + a.​

Answer 1

Answer:

(a+b+c)³=[(a+b)+c]³=(a+b)³+3(a+b)²c+3(a+b)c²+c³

(a+b+c)3=(a³+3a²b+3ab²+b³)+3(a²+2ab+b²)c+3(a+b)c2+c³

(a+b+c)3=a³+b³+c³+3a²b+3a²c+3ab²+3b²c+3ac²+3bc²+6abc

(a+b+c)³=(a³+b³+c³)+(3a²b+3a²c+3abc)+(3ab²+3b²c+3abc)+(3ac²+3bc²+3abc)−3abc

The next three expansions are for those who refrain from exponents in expansions. (lol)

(a+b+c)³=(a³+b³+c³)+3a(ab+ac+bc)+ 3b(ab+bc+ac) +3c(ac+bc+ab)−3abc

(a+b+c)³=(a³+b³+c³)+3(a+b+c)(ab+ac+bc)−3abc

(a+b+c)3=(a3+b3+c3)+3[(a+b+c)(ab+ac+bc)−abc]

Hope I could be helpful to you. Thank you and have a glorious day applying these.

All the very best!! :-)

Answer 2

Answer:

(a+b+c)³=(a³+b³+c³)+3a(ab+ac+bc)+ 3b(ab+bc+ac) +3c(ac+bc+ab)−3abc

(a+b+c)³=(a³+b³+c³)+3(a+b+c)(ab+ac+bc)−3abc

(a+b+c)3=(a3+b3+c3)+3[(a+b+c)(ab+ac+bc)−abc]

Step-by-step explanation: