Question

guys fast do it
I will mark brinliest to 1$t answer
question 19 ​

Answer 1

Solution:

(a+b)^3 +(b+c)^3 +(c+a)^3 - 3(a+b)(b+c)(c+a)

= a^3+b^3 + 3(a^2)b + 3a(b^2) + b^3 + c^3 + 3(b^2)c + 3b(c^2) + a^3 + c^3 + 3(a^2)c + 3a(c^2) - 3(b^2)c - 3b(c^2) - 3abc - 3a(c^2) - 3a(b^2) - 3abc - 3(a^2)b - 3(a^2)c

= 2a^3 + 2b^3 + 2c^3 + 6abc

=2(a^3 + b^3 + c^3 + 3abc)

Hope it Helped all...:-)

Answer 2

Answer:

Step-by-step explanation:

Answer:

Step-by-step explanation:

(a+b)³+(b+c)³+(c+a)³-3(a+b)(b+c)(c+a)=2(a³+b³+c³)-3abc

a³+b³+3ab(a+b)+b³+c³+3bc(b+c)+c³+a³+3ca(c+a)-3(a+b)(b+c)(c+a)=2(a³+b³+c³)-3abc

2a³+2b³+2c³+3ab(a+b)+3bc(b+c)+3ca(c+a)-3(a+b)(b+c)(c+a)=2(a³+b³+c³)-3abc

2a³+2b³+2c³+[3abc(a+b)(b+c)(c+a)-3(a+b)(b+c)(c+a)]=2(a³+b³+c³)-3abc

2a³+2b³+2c³+[3abc{3(a+b)(b+c)(c+a)}-{3(a+b)(b+c)(c+a)}]=2(a³+b³+c³)-3abc

2a³+2b³+2c³+3abc=2(a³+b³+c³)-3abc

2(a³+b³+c³)+3abc=2(a³+b³+c³)-3abc

PROVED