prove that : (a+b)^3 +(b+c)^3+(c+a)^3-3(a+b)(b+c)(c+a) = 2(a^3+b^3+c^3-3abc)
Answers
Answered by
6
hope helpful budy.......
☺☺☺ if helpful mark me as brainliest
☺☺☺ if helpful mark me as brainliest
Attachments:
rohitkumargupta:
hehe
ok i will.... ranjan mahasay jii
hehe
stop yar
ohh i am just joking i am not angry dont need to sorry
he he
oka
Answered by
11
HELLO DEAR,
now expand ,
(a+b)³ = a³+b³+3ab(a+b)
(b+c)³ = b³+c³+3bc(b+c)
(c+a)³ = c³+a³+3ca(c+a)
---according to question:---
=> a³+b³+3ab(a+b) + b³+c³+3bc(b+c) + c³+a³+3ca(c+a) -3[(a+b)(b+c)(c+a)]
=> 2a³+2b³+2c² + 3[ab(a+b) + ca(c+a) + bc(b+c)]
- 3[(a+b)(b+c)(c+a)]
=> 2(a³+b³+c³) + 3 [ a²b+ b²a +c²a +a²c +b²c +c²b] - 3[ (ab+ac+b²+bc)(c+a)]
=>2(a³+b³+c³) + 3 [ a²b+ b²a +c²a +a²c +b²c +c²b] - 3[abc + ac² +b²c + bc² + a²b + a²c +ab² +abc]
=> 2(a³+b³+c³) + 3a²b+3b²a +3c²a -3a²c +3b²c +3c²b - 3abc - 3ac² -3b²c - 3bc² - 3a²b - 3a²c -3ab² - 3abc
=> 2(a³+b³+c³) -6abc
=>2(a³+b³+c³-3abc)
I HOPE ITS HELP YOU DEAR,
THANKS
now expand ,
(a+b)³ = a³+b³+3ab(a+b)
(b+c)³ = b³+c³+3bc(b+c)
(c+a)³ = c³+a³+3ca(c+a)
---according to question:---
=> a³+b³+3ab(a+b) + b³+c³+3bc(b+c) + c³+a³+3ca(c+a) -3[(a+b)(b+c)(c+a)]
=> 2a³+2b³+2c² + 3[ab(a+b) + ca(c+a) + bc(b+c)]
- 3[(a+b)(b+c)(c+a)]
=> 2(a³+b³+c³) + 3 [ a²b+ b²a +c²a +a²c +b²c +c²b] - 3[ (ab+ac+b²+bc)(c+a)]
=>2(a³+b³+c³) + 3 [ a²b+ b²a +c²a +a²c +b²c +c²b] - 3[abc + ac² +b²c + bc² + a²b + a²c +ab² +abc]
=> 2(a³+b³+c³) + 3a²b+3b²a +3c²a -3a²c +3b²c +3c²b - 3abc - 3ac² -3b²c - 3bc² - 3a²b - 3a²c -3ab² - 3abc
=> 2(a³+b³+c³) -6abc
=>2(a³+b³+c³-3abc)
I HOPE ITS HELP YOU DEAR,
THANKS
Similar questions